Question

A ball player catches a ball 3.33 s after throwing it vertically upward. With what speed did he throw it?

Answer 1

Answer:

Speed = 16.317 m/s

Height = 13.58 m

Explanation:

If he catches the ball 3.33 s after throwing it, we can say that it takes half the time to get to the highest point.

So, it takes 3.33 s/2 = 1.665 s to reach the highest point and the final velocity at that point was 0 m/s

Now, we can use the following equation:

$v_f=v_0_{}+gt$

Where vf is the final velocity, v0 is the initial velocity, g is the acceleration in this case and it is equal to -9.8 m/s² and t is the time.

So, replacing the values, we get:

$0=v_0-9.8(1.665)$

Solving for v0, we get:

$\begin{gathered} 0=v_0-16.317 \\ 0+16.317=v_0-16.317+16.317 \\ 16.317m/s=v_0 \end{gathered}$

Therefore, the answer is 16.317 m/s

Now, the maximum height of the ball can be found using the following equation:

$y=y_0+v_0t+\frac{1}{2}at^2$

Where y0 is the initial height and y is the final height. So, replacing the values, we get:

$\begin{gathered} y=0+16.317(1.665)-\frac{1}{2}(9.8)(1.665)^2 \\ y=13.58m \end{gathered}$

Therefore, the maximum height of the ball was 13.58 m