Chapter 3, Problem 69QAP

To determine

(a)

The time taken by a ball released from an airplane to hit the ground.

Answer to Problem 69QAP

The ball hits the ground after 9.24 s.

Explanation of Solution

Given:

The speed of the airplane when the ball is released

  $v_0=35.3\text{ m/s}$

Angle made by the plane to the horizontal

  $\theta =30.0°$

Height of the airplane above the ground when the ball is released $\Delta y=-255\text{ m}$

Formula used:

The time of flight of the ball is determined using the equation for the vertical motion of the ball.

  $\Delta y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$......(1)

Here, $\Delta y$ is the total vertical displacement of the ball, $v_{0y}$ is the initial vertical component of the ball's velocity, $a_y$ is the acceleration acting on the ball in the vertical direction and t is the time taken by the ball to make the vertical displacement of $\Delta y$.

Calculation:

When the airplane releases the ball, the ball has the velocity of the airplane. Its speed is 35.3 m/s and it is released at an angle 30.0^o to the horizontal.

Assume the origin to be located at the point where the ball is released. With the x axis parallel to the ground and the + y axis directed upwards.

The ball travels a parabolic path and lands on the ground at point B. Its vertical displacement when it lands on the ground, is equal to $\Delta y$

This is shown in the diagram below.

Calculate the vertical component of the ball's velocity.

  $\begin{array}{c}{v}_{0y}=v_0\sin \theta \\ =\left(35.3\text{ m/s}\right)\left(\sin 30.0°\right)\\ =17.65\text{ m/s}\end{array}$

The ball falls under the action of the gravitational force. Hence the acceleration acting on the ball in the vertical direction is the acceleration of free fall.

  $a_y=g=-9.80{\text{ m/s}}^2$

  

In equation (1), substitute $\left(-255.0\text{ m}\right)$ for $\Delta y$, $\left(17.65\text{ m/s}\right)$ for $v_{0y}$ and $\left(-9.80{\text{ m/s}}^2\right)$ for $a_y$ and write a quadratic equation in t.

  $\left(-255.0\text{ m}\right)=\left(17.65\text{ m/s}\right)t+\frac{1}{2}\left(-9.80{\text{ m/s}}^2\right)t^2$

  $\left(4.90{\text{ m/s}}^2\right)t^2-\left(17.65\text{ m/s}\right)t+\left(-255.0\text{ m}\right)=0$

Solve the quadratic equation to determine t.

  $\begin{array}{c}t=\frac{-\left(-17.65\text{ m/s}\right)\pm \sqrt{{\left(-17.65\text{ m/s}\right)}^2-4\left(4.90{\text{ m/s}}^2\right)\left(-255.0\text{ m}\right)}}{2\left(4.90{\text{ m/s}}^2\right)}\\ =\frac{\left(17.65\text{ m/s}\right)\pm \left(72.87\text{ m/s}\right)}{\left(9.80{\text{ m/s}}^2\right)}\end{array}$

Taking the positive root,

  $t=9.24\text{ s}$

Conclusion:

Thus, the ball hits the ground after 9.24 s.

To determine

(b)

The maximum height of the ball from the ground.

Answer to Problem 69QAP

The ball reaches a maximum height of 270.9 m from the ground.

Explanation of Solution

Given:

The speed of the airplane when the ball is released

  $v_0=35.3\text{ m/s}$

Angle made by the plane to the horizontal

  $\theta =30.0°$

Height of the airplane above the ground when the ball is released $\Delta y=-255\text{ m}$

Formula used:

The maximum height reached by the ball can be calculated using the equation of motion,

  $v_y^2=v_{0y}^2+2a_y\left(y-y_0\right)$......(2)

Here, v_y is vertical component of the ball's velocity at the position y, $v_{0y}$ is the initial vertical component of the ball's velocity, $a_y$ is the acceleration acting on the ball in the vertical direction and y₀ is the y coordinate of the point where the ball is launched.

The maximum height h reached by the ball, when measured from the ground is given by,

  $h=y-\Delta y$

Calculation:

The vertical component of the ball's velocity reduces as it moves up, due to the action of the gravitational force. When the vertical component reaches a value zero, the ball can no longer make an upward displacement, hence after this point it starts its motion in the downward direction.

Therefore, at maximum height,

  $v_y=0\text{ m/s}$

In equation (2) substitute 0 m/s for v_y, $\left(17.65\text{ m/s}\right)$ for $v_{0y}$, $\left(-9.80{\text{ m/s}}^2\right)$ for $a_y$ and 0 m for y₀.

  $\begin{array}{c}{v}_y^2=v_{0y}^2+2a_y\left(y-y_0\right)\\ {\left(0\text{ m/s}\right)}^2={\left(17.65\text{ m/s}\right)}^2+2\left(-9.80{\text{ m/s}}^2\right)\left[y-\left(0\text{ m}\right)\right]\\ y=-\frac{{\left(17.65\text{ m/s}\right)}^2}{2\left(-9.80{\text{ m/s}}^2\right)}\\ =15.89\text{ m}\end{array}$

This point is 15.89 m above the point of projection. Therefore, its height from the ground is given by,

  $\begin{array}{c}h=y-\Delta y\\ =\left(15.89\text{ m}\right)-\left(-255.0\text{ m}\right)\\ =270.2\text{ m}\end{array}$

Conclusion:

Thus, the ball reaches a maximum height of 270.9 m from the ground.

To determine

(c)

The horizontal distance traveled by the ball from the point of release to the ground.

Answer to Problem 69QAP

The ball travels a horizontal distance of 282.5 m.

Explanation of Solution

Given:

The speed of the airplane when the ball is released

  $v_0=35.3\text{ m/s}$

Angle made by the plane to the horizontal

  $\theta =30.0°$

Height of the airplane above the ground when the ball is released $\Delta y=-255\text{ m}$

Time of flight of the ball

  $t=9.24\text{ s}$

Formula used:

The horizontal distance traveled by the ball is calculated using the equation

  $\Delta x=v_{0x}t+\frac{1}{2}{a}_x{t}^2$......(3)

Here, $\Delta x$ is the horizontal displacement of the ball, $v_{0x}$ is the horizontal component of the ball's velocity, $a_x$ is the acceleration along the x direction and t is the time of flight of the ball.

Calculation:

The ball makes a vertical displacement of $\Delta y$ when it reaches the ground and it takes a time t to reach the ground. At the same time, the ball also makes a horizontal displacement $\Delta x$. No force acts on the ball in the horizontal direction, hence the acceleration $a_x=0{\text{ m/s}}^2$.

Calculate the horizontal component of the ball's velocity.

  $\begin{array}{c}{v}_{0x}=v_0\cos \theta \\ =\left(35.3\text{ m/s}\right)\left(\cos 30.0°\right)\\ =30.57\text{ m/s}\end{array}$

Substitute the values of v_0x, a_x and t in equation (3) and solve for $\Delta x$.

  $\begin{array}{c}\Delta x=v_{0x}t+\frac{1}{2}{a}_x{t}^2\\ =\left(30.57\text{ m/s}\right)\left(9.24\text{ s}\right)+\frac{1}{2}\left(0{\text{ m/s}}^2\right){\left(9.24\text{ s}\right)}^2\\ =282.5\text{ m}\end{array}$

Conclusion:

Thus, the ball travels a horizontal distance of 282.5 m.