Question

A ball is thrown toward a cliff of height h with a speed of 30 m/s and an angle of 60° above horizontal. It lands on the edge of the cliff 4.0 s later.
a. How high is the cliff?
b. What was the maximum height of the ball?
c. What is the ball’s impact speed?

Short answer

a) The height of the cliff is 25.5 m.

b) The maximum height the ball reached is 34.43m.
c) The impact velocity of the ball is 20m/s

Step-by-step solution

Write the given information

The initial speed of the ball, u= 30m/s
The angle with the horizontal, θ = 60⁰
The ball hit the cliff after time, t= 4sec

 Step 2: a) To determine the height of the cliff

Let the height of the cliff is H
The horizontal component of the velocity is
$$\begin{gathered} v_x=v\cos 60 \\ {v}_x=15m/sec \end{gathered}$$

The vertical component of the velocity is
$$\begin{gathered} v_y=v\sin (60°) \\ {v}_y=25.9m/s \end{gathered}$$

By using the equation of motion

role="math" localid="1650529655143" $$\begin{gathered} H=v_{v}t+\frac{1}{2}g{t}^2 \\ H=25.9\left(4\right)+\frac{1}{2}(-9.8){\left(4\right)}^2 \\ H=25.5m \end{gathered}$$
Thus, the height of the cliff is 25.5 m.

b) To determine the maximum height of the ball.

At the maximum height, the velocity of the ball would be zero
Using the equation of motion
$$\begin{gathered} v^2-{v_y}^2=2aH \\ 0-{(25.9)}^2=2(-9.8)H_{max} \\ {H}_{max}=34.43m \end{gathered}$$

Thus, the maximum height the ball reached is 34.43m.

c) To determine the impact speed of the ball

The impact speed of the ball is the speed at which the ball hits the cliff.
The vector component of velocity at that point is shown below

The difference between the maximum height of the ball and the height of the cliff is given by

$$\begin{gathered} y=H_{max}-H \\ y=34.43m-25.5m \\ y=8.9m \end{gathered}$$
The time taken by the ball to cover this distance,
$$\begin{gathered} t=\sqrt{\frac{2y}{g}} \\ t=\sqrt{\frac{2(8.9)}{9.8}} \\ t=1.35s \end{gathered}$$

Now, the vertical velocity of the ball is given by,

$$\begin{gathered} v_y=u+gt \\ {v}_y=0+\left(9.8\right)(1.35) \\ {v}_y=13.2m/s \end{gathered}$$

The impact velocity of the ball is given by
$$\begin{gathered} {v_r}^2={v_y}^2+{\left(v\cos 60°\right)}^2 \\ {v_r}^2={(13.2)}^2+{\left(25.9\right)}^2{v}_r=20m/sec \end{gathered}$$

The impact velocity of the ball is 20m/s