Question

A ball player catches a ball 3.2 s after throwing it vertically upward.With what speed did he throw it? What height did it reach?Express your answers using two significant figures.

Answer 1

Since the time of flight of the ball is 3.2 seconds, that means for 1.6 seconds the ball was going upwards, then it reached the maximum height (where the velocity is zero), then went 1.6 seconds downwards, until it reaches the hand of the player again.

Using the formula below, we can find the initial velocity of the ball:

`V=V_0+a\cdot t`

Where V is the final velocity after t seconds, V0 is the initial velocity and 'a' is the acceleration.

Using V = 0, t = 1.6 s and a = -9.8 (gravity's acceleration), we have:

`\begin{gathered} 0=V_0-9.8\cdot1.6\\ \\ V_0-15.68=0\\ \\ V_0=15.68\text{ m/s} \end{gathered}`

Rounding the answer to two significant figures, we have an initial velocity of 16 m/s.

To find the maximum height, we can use the formula below:

`\begin{gathered} \Delta S=V_0t+(at^2)/(2)\\ \\ \Delta S=15.68\cdot1.6-(9.8\cdot1.6^2)/(2)\\ \\ \Delta S=25.088-12.544\\ \\ \Delta S=12.544\text{ m} \end{gathered}`

Rounding the answer to two significant figures, the maximum height is 13 m.