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Problem 6-8: An integrated problem involving circular motion, momentum, and projectile motion

Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

Solution:

Part A

From the given problem, we are given the following values: v=35.0 m/sv=35.0\ \text{m/s} and r=1.05 mr=1.05\ \text{m}. We are required to solve for the angular velocity ω\omega.

The linear velocity, v v and the angular velocity, ω \omega are related by the equation

v=rω or ω=vrv=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into the formula, we can directly solve for the value of the angular velocity. That is,

ω=vrω=35.0 m/s1.05 mω=33.3333 rad/secω=33.3 rad/s  (Answer)\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{35.0\ \text{m/s}}{1.05\ \text{m}} \\ \\ \omega & = 33.3333\ \text{rad/sec} \\ \\ \omega & = 33.3 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this part of the problem, we are going to use Newton’s second law of motion in term of linear momentum which states that the net external force equals the change in momentum of a system divided by the time over which it changes. That is

Fnet=ΔpΔt=m(vf−vi)tF_{net} = \frac{\Delta p}{\Delta t} = \frac{m\left( v_f - v_i \right)}{t}

For this problem, we are given the following values: m=0.500 kgm=0.500\ \text{kg}, t=20.0×10−3 st=20.0\times 10^{-3} \ \text{s}, vf=20.0 m/sv_{f}=20.0\ \text{m/s}, and vi=0v_{i}=0. Substituting all these values into the equation, we can solve directly for the value of the net external force.

Fnet=(0.500 kg)(20.0 m/s−0 m/s)20.0×10−3 sFnet=500 N  (Answer)\begin{align*} F_{net} & = \frac{\left( 0.500\ \text{kg} \right)\left( 20.0\ \text{m/s}-0\ \text{m/s} \right)}{20.0\times 10^{-3}\ \text{s}} \\ \\ F_{net} & = 500\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

This is a problem on projectile motion. In this particular case, we are solving for the range of the projectile. The formula for the range of a projectile is

R=v02sin⁡2θgR=\frac{v_{0}^2 \sin 2\theta}{g}

We are asked to solve for the maximum range, and we know that the maximum range happens when the angle θ\theta is 45∘45^\circ .

R=(20.0 m/s)2sin⁡(2(45∘))9.81 m/s2R=40.7747 mR=40.8 m  (Answer)\begin{align*} R & = \frac{\left( 20.0\ \text{m/s} \right)^{2} \sin \left( 2\left( 45^\circ \right) \right)}{9.81 \ \text{m/s}^2} \\ \\ R & = 40.7747\ \text{m} \\ \\ R & = 40.8 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 3 Problem 37

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The baseline is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose service line is 6.40 m from the net?

Solution:

Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.

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College Physics by Openstax Chapter 3 Problem 36

The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

Solution:

We are required to solve for the maximum distance. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

  • The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
  • The jumper is on level ground, and the motion started from the ground.

The formula for range is

R=vo2sin⁡2θog\text{R}=\frac{\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}}}{\text{g}}

Since we are already given the necessary details, we can now solve for the range.

R=(9.5 m/s)2 sin⁡90∘9.81 m/s2R=9.20 m  (Answer)\begin{align*} \text{R}&=\frac{\left(9.5\:\text{m/s}\right)^2\:\sin 90^{\circ} }{9.81\:\text{m/s}^2}\\ \text{R}&=9.20\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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College Physics by Openstax Chapter 3 Problem 35

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Solution:

We are required to solve for the distance in a standing broad jump. To do this, we can use the formula for the range of a projectile motion. However, we need the following assumptions:

  • The jumper leaves the ground in a 45° angle from the horizontal, for maximum horizontal displacement.
  • The jumper is on level ground.

The formula for the range is

R=vo2 sin⁡2θog\text{R}=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}}}{\text{g}}

To find the initial velocity of the jump, vo, we shall use the kinematic formula from the crouch position to the time the person leaves the ground.

vf2=vo2+2ax\text{v}_{\text{f}}^2=\text{v}_{\text{o}}^2+2\text{ax}

In this case, the final velocity will be the initial velocity of the jump.

vf=(0 m/s)2+2(1.25×9.81 m/s2)(0.600 m)=3.84 m/s\begin{align*} \text{v}_{\text{f}}=\sqrt{\left(0\:\text{m/s}\right)^2+2\left(1.25\times 9.81\:\text{m/s}^2\right)\left(0.600\:\text{m}\right)}=3.84\:\text{m/s} \end{align*}

So, the initial velocity of the flight is 3.84 m/s. We can now use the formula for range.

R=vo2 sin⁡2θogR=(3.84 m/s)2 sin⁡(2×45∘)9.81 m/s2R=1.50 m  (Answer)\begin{align*} \text{R}&=\frac{\text{v}_{\text{o}}^2\:\sin 2\theta_{\text{o}}}{\text{g}} \\ \text{R}&=\frac{\left(3.84\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}\\ \text{R}&=1.50\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 3 Problem 34

An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?

Solution:

Consider the following illustration:

An arrow shot at a height of 1.5 m towards a cliff of height H

Part A

We are required to solve for the value of H. We shall use the formula

Δy=v0yt+12at2\Delta \text{y}=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

or, we can also write the formula as

y−y0=v0yt+12at2 \text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

Substituting the given values, we have

y−y0=v0yt+12at2H−1.5 m=(30 m/s sin⁡60∘)(4.0 s)+12(−9.81 m/s2)(4.0 s)2H−1.5 m=25.44 mH=25.44 m+1.5 mH=26.94 m  (Answer)\begin{align*} \text{y}-\text{y}_0 &=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2 \\ \text{H}-1.5\:\text{m} & =\left(30\:\text{m/s}\:\sin 60^{\circ} \right)\left(4.0\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right)^2 \\ \text{H}-1.5\:\text{m} & =25.44\:\text{m} \\ \text{H} & =25.44\:\text{m}+1.5\:\text{m} \\ \text{H} & =26.94\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Part B

The maximum height of the projectile is given by the formula

Δy=v0y22g\Delta \text{y}=\frac{\text{v}_{\text{0y}}^2}{2\text{g}}

or the formula can be written as

ymax−y0=v0y2−2g\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}

Therefore, we have

ymax−y0=v0y2−2gymax=v0y2−2g+y0ymax=((30 m/s)sin⁡60∘)2−2(−9.81 m/s2)+1.5 mymax=34.40 m+1.5 mymax=35.90 m  (Answer)\begin{align*} \text{y}_{\text{max}}-\text{y}_{\text{0}} & =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}} \\ \text{y}_{\text{max}}& =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}+\text{y}_{\text{0}} \\ \text{y}_{\text{max}}&=\frac{\left(\left(30\:\text{m/s}\right)\sin 60^{\circ} \right)^2}{-2\left(-9.81\:\text{m/s}^2\right)}+1.5\:\text{m} \\ \text{y}_{\text{max}}&=34.40\:\text{m}+1.5\:\text{m} \\ \text{y}_{\text{max}} & =35.90\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Part C

To solve for the final speed, we need the vertical and horizontal components when the arrow hits the cliff.

We are going to use the formula for acceleration along the vertical to solve for the final speed in the vertical direction. That is

a=vy−v0yt\text{a}=\frac{\text{v}_{\text{y}}-\text{v}_{\text{0y}}}{\text{t}}

Solving for vfy in terms of the other variables, we have

vy=v0y+atvy=(30 m/s)sin⁡60∘+(−9.81 m/s2)(4.0 s)vy=−13.25 m/s\begin{align*} \text{v}_{\text{y}}&=\text{v}_{\text{0y}}+\text{at}\\ \text{v}_{\text{y}}&=\left(30\:\text{m/s}\right)\sin 60^{\circ} +\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right) \\ \text{v}_{\text{y}}&=-13.25\:\text{m/s} \end{align*}

Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is

vx=v0x=(30 m/s)cos⁡60∘vx=15 m/s\begin{align*} \text{v}_{\text{x}}&=\text{v}_{\text{0x}}=\left(30\:\text{m/s}\right)\cos 60^{\circ} \\ \text{v}_{\text{x}}&=15\:\text{m/s} \end{align*}

Therefore, the final speed is

v=vy2+vx2v=(−13.25 m/s)2+(15 m/s)2v=20.01 m/s  (Answer)\begin{align*} \text{v}&=\sqrt{\text{v}_{\text{y}}^2+\text{v}_{\text{x}}^2} \\ \text{v}&=\sqrt{\left(-13.25\:\text{m/s}\right)^2+\left(15\:\text{m/s}\right)^2}\\ \text{v}&=20.01\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 3 Problem 33

The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

Solution:

Part A

We are given the range of the projectile motion. The range is 32.0 km. We also know that for the projectile to reach its maximum distance, it should be fired at 45°. So from the formula of range,

R=v02 sin⁡2θ0g\displaystyle \text{R}=\frac{\text{v}_0^2\:\sin 2\theta _0}{\text{g}}

we can say that sin⁡2θ0=sin⁡(2×45∘)=sin⁡90∘=1 \sin 2\theta _0=\sin \left(2\times 45^{\circ} \right)=\sin 90^{\circ} =1. So, we have

R=v02g\displaystyle \text{R}=\frac{\text{v}_0^2}{\text{g}}

We can solve for v0 in terms of the other variables. That is

v0=gR\displaystyle \text{v}_0=\sqrt{\text{gR}}

Substituting the given values, we have

v0=(9.81 m/s2)(32×103 m)v0=560.29 m/s  (Answer)\begin{align*} \displaystyle \text{v}_0 & =\sqrt{\left(9.81\:\text{m/s}^2\right)\left(32\times 10^3\:\text{m}\right)} \\ \displaystyle \text{v}_0 & =560.29\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part B

We are solving for the maximum height here, which happened at the mid-flight of the projectile. The vertical velocity at this point is zero. Considering all this, the formula for the maximum height is derived to be

hmax=v0y22g\displaystyle \text{h}_{\text{max}}=\frac{\text{v}_{0_y}^2}{2\text{g}}

The initial vertical velocity, v0y, is calculated as

v0y=v0sin⁡θ0=(560.29 m/s)sin⁡45∘=396.18 m/s\begin{align*} \text{v}_{\text{0y}} & =\text{v}_0\sin \theta _0 \\ & =\left(560.29\:\text{m/s}\right)\sin 45^{\circ} \\ & =396.18\:\text{m/s} \end{align*}

Therefore, the maximum height is

hmax=(396.18 m/s)22(9.81 m/s2)hmax=8000 m  (Answer)\begin{align*} \text{h}_{\text{max}} & =\frac{\left(396.18\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\ \text{h}_{\text{max}} & =8000\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part C

Consider the following figure

A right triangle is formed with the legs, the horizontal distance and the radius of the earth, and the hypotenuse is the sum of the radius of the earth and the distance d, which is the unknown in this problem. Using Pythagorean Theorem, and converting all units to meters, we have

R2+(32.0×103 m)2=(R+d)2(6.37×106 m)2+(32.0×103 m)2=(6.37×106+d)2d=(6.37× 106 )2+(32.0×103 )2−6.37× 106d=80.37 m  (Answer)\begin{align*} \text{R}^2+\left(32.0\times 10^3\:\text{m}\right)^2 & =\left(\text{R}+\text{d}\right)^2 \\ \left(6.37\times 10^6\:\text{m}\right)^2+\left(32.0\times 10^3\:\text{m}\right)^2 & =\left(6.37\times 10^6+\text{d}\right)^2 \\ \text{d} & =\sqrt{\left(6.37\times \:10^6\:\right)^2+\left(32.0\times 10^3\:\right)^2}-6.37\times \:10^6 \\ \text{d} & =80.37\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

This error is not significant because it is only about 1% of the maximum height computed in Part B.

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College Physics by Openstax Chapter 3 Problem 32

Verify the ranges shown for the projectiles in Figure 3.40(b) for an initial velocity of 50 m/s at the given initial angles.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial angles. To do this, we shall use the formula

R=v02sin⁡2θ0g\displaystyle \text{R}=\frac{\text{v}_{\text{0}}^2 \sin 2\theta _{\text{0}}}{\text{g}}

When the initial angle is 15°, the range is

R=(50 m/s)2 sin⁡(2×15∘)9.81 m/s2=127.42 m\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 15^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

When the initial angle is 45°, the range is

R=(50 m/s)2 sin⁡(2×45∘)9.81 m/s2=254.84 m\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

When the initial angle is 75°, the range is

R=(50 m/s)2 sin⁡(2×75∘)9.81 m/s2=127.42 m\displaystyle \text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 75^{\circ} \right)}{9.81\:\text{m/s}^2}=127.42\:\text{m}

Based on the result of the calculations, we can say that the numbers in the figure are verified. The very small differences are only due to round-off errors.

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College Physics by Openstax Chapter 3 Problem 31

Verify the ranges for the projectiles in Figure 3.40(a) for θ=45º and the given initial velocities.

Solution:

To verify the given values in the figure, we need to solve for individual ranges for the given initial velocities. To do this, we shall use the formula

R=v02 sin⁡2θ0g\text{R}=\frac{\text{v}_{\text{0}}^2\:\sin 2\theta _{\text{0}}}{\text{g}}

When the initial velocity is 30 m/s, the range is

R=(30 m/s)2 sin⁡(2×45∘)9.81 m/s2=91.74 m\text{R}=\frac{\left(30\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=91.74\:\text{m}

When the initial velocity is 40 m/s, the range is

R=(40 m/s)2 sin⁡(2×45∘)9.81 m/s2=163.10 m\text{R}=\frac{\left(40\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=163.10\:\text{m}

When the initial velocity is 50 m/s, the range is

R=(50 m/s)2 sin⁡(2×45∘)9.81 m/s2=254.84 m\text{R}=\frac{\left(50\:\text{m/s}\right)^2\:\sin \left(2\times 45^{\circ} \right)}{9.81\:\text{m/s}^2}=254.84\:\text{m}

Based on the results, we can say that the ranges are approximately equal. The differences are only due to round-off errors.

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College Physics by Openstax Chapter 3 Problem 30

A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

Solution:

To illustrate the problem, consider the following figure:

A player passes the ball 7 meters across the field with an initial velocity of 12 m/s

Part A

We are given the 7-meter range, R, and the initial velocity, vo, of the projectile. We have R=7.0 m, and vo=12.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

R=vo2sin⁡2θog\text{R}=\frac{\text{v}^{2}_{\text{o}}\sin 2\theta _{\text{o}}}{g}

Solving for θo in terms of the other variables, we have

gR=vo2sin⁡2θosin⁡2θo=gRvo22θo=sin⁡−1(gRvo2)θo=12sin⁡−1(gRvo2)\begin{align*} \text{gR} & =\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}} \\ \sin 2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\ 2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\ \theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\ \end{align*}

Substituting the given values, we have

θo=12sin⁡−1[(9.81m/s2)(7.0m)(12.0m/s)2]θo=14.2∘(Answer)\begin{align*} \theta _\text{o} & =\frac{1}{2} \sin ^{-1}\left[\frac{\left(9.81\text{m/s}^2\right)\left(7.0\text{m}\right)}{\left(12.0\text{m/s}\right)^2}\right] \\ \theta _\text{o} & =14.2^{\circ} \qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

Part B

The other angle that would give the same range is actually the complement of the solved angle in Part A. The other angle,

θo′=90∘−14.24∘=75.8∘(Answer)\theta _o'=90^{\circ} -14.24^{\circ} =75.8^{\circ} \qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\

This angle is not used as often, because the time of flight will be longer. In rugby that means the defense would have a greater time to get into position to knock down or intercept the pass that has the larger angle of release.

Part C

We can use the x-component of the motion to solve for the time of flight.

Δx=vxt\Delta \text{x}=\text{v}_\text{x}\text{t}

We need the horizontal component of the velocity. We should be able to solve for the component since we are already given the initial velocity and the angle.

vx=(12 m/s)cos⁡14.24∘=11.63 m/s\text{v}_{\text{x}}=\left(12\:\text{m/s}\right)\cos 14.24^{\circ} =11.63\:\text{m/s}

Therefore, the total time of flight is

t=Δxvxt=7.0 m11.63 m/st=0.60 s(Answer)\begin{align*} \text{t} & =\frac{\Delta \text{x}}{\text{v}_{\text{x}}} \\ \text{t} & =\frac{7.0\:\text{m}}{11.63\:\text{m/s}} \\ \text{t} & =0.60\:\text{s} \qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}
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College Physics by Openstax Chapter 3 Problem 29

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

Solution:

To illustrate the problem, consider the following figure:

The archer and the target at 75 meter range

Part A

We are given the range of 75-meter range, R, and the initial velocity, vo, of the projectile. We have R=75.0 m, and vo=35.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

R=vo2 sin⁡2θog\text{R}=\frac{\text{v}^{2}_{\text{o}}\:\sin 2\theta _{\text{o}}}{g}

Solving for θo in terms of the other variables, we have

gR=vo2 sin⁡2θosin⁡ 2θo=gRvo22θo=sin⁡−1(gRvo2)θo=12sin⁡−1(gRvo2)θo=12sin⁡−1[(9.81 m/s2)(75.0 m)(35.0 m/s)2]θo=18.46∘  (Answer)\begin{align*} \text{gR} & =\text{v}_{\text{o}}^2\:\sin 2\theta _{\text{o}} \\ \sin \:2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\ 2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\ \theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\ \theta _o & =\frac{1}{2}\sin ^{-1}\left[\frac{\left(9.81\:\text{m/s}^2\right)\left(75.0\:\text{m}\right)}{\left(35.0\:\text{m/s}\right)^2}\right] \\ \theta _o & =18.46^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

We know that halfway, the maximum height of the projectile occurs. Also at this instant, the vertical velocity is zero. We can solve for the maximum height and compare it with the given height of 3.50 meters.

The maximum height can be computed using the formula

hmax=voy22g\text{h}_{\text{max}}=\frac{\text{v}_{\text{oy}}^2}{2\text{g}}

To compute for the maximum height, we need the initial vertical velocity, voy. Since we know the magnitude and direction of the initial velocity, we have

voy=(35.0 m/s)sin⁡18.46∘voy=11.08 m/s\begin{align*} \text{v}_{\text{oy}} & =\left(35.0\:\text{m/s}\right)\sin 18.46^{\circ} \\ \text{v}_{\text{oy}} & =11.08\:\text{m/s} \end{align*}

Therefore, the maximum height is

hmax⁡=(11.08 m/s)22(9.81 m/s2)hmax⁡=6.26 m  (Answer)\begin{align*} \text{h}_{\max } & =\frac{\left(11.08\:\text{m/s}\right)^2}{2\left(9.81\:\text{m/s}^2\right)} \\ \text{h}_{\max } & =6.26\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

We have known that the path of the arrow is above the branch of the tree. Therefore, the arrow will go through.

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