Category Archives: Sciences

Part A: To convert the given mass in kilogram to newton force, we simply need to multiply by the acceleration due to gravity of 9.81 m/s². We need to take into account that $1\:\text{kg m/s}^2\:=1\:\text{N}$ .

\begin {aligned} 8\:\text{kg} & =8\:\text{kg}\times 9.81\:\text{m/s}^2 \\ &=78.48\:\text{N} \end {aligned}

Part B: Using the same principle from Part A, we have

\begin {aligned} 0.04\:\text{kg}&=0.04\:\text{kg}\times 9.81\:\text{m/s}^2\\ &=0.3924\:\text{N} \end {aligned}

Part C: So, we are given 760 Mg (megagram). We know that 1 Mg is equivalent to 1000 kg. Therefore, 760 Mg is equal to 760,000 kg. Therefore, we have

\begin {aligned} 760\:000\:\text{kg}&=760\:000\:\text{kg}\times 9.81\:\text{m/s}^2\\ &=7\:455\:600\:\text{N} \end{aligned}

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Physics

College Physics by Openstax

Physics for Scientists and Engineers: A Strategic Approach Third Edition by Randall D. Knight

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics 3rd Edition by Randall D. Knight

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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

\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.

\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

\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.

\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.

\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.

\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

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

or, we can also write the formula as

\text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2

Substituting the given values, we have

\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

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

or the formula can be written as

\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}

Therefore, we have

\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

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

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

\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

\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

\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,

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

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

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

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

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

Substituting the given values, we have

\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

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

The initial vertical velocity, v_0y, is calculated as

\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

\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

\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

\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

\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

\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

\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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