Vector Addition and Subtraction: Graphical Methods
Vector Addition and Subtraction: Analytical Methods
Projectile Motion
Addition of Velocities
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Vector Addition and Subtraction: Graphical Methods
Vector Addition and Subtraction: Analytical Methods
Projectile Motion
Addition of Velocities
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Solution:
Note: The publication of the solution to this problem is on its way. Sorry for the inconvenience.
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 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*}
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 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*}
Solution:
Consider the following illustration:
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*}
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*}
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 vfy 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*}
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, v0y, 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.
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.
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
\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
\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
\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
\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.
Solution:
To illustrate the problem, consider the following figure:
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
\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
\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
\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,
\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.
\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.
\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
\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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