Question

A ball player catches the ball 3 seconds after throwing it straight up in the air. What speed did he throw it and how high did it go

Answer 1

Given:

A player catches the ball 3 seconds after throwing it straight up in air.

Let's find the speed and the maximum height of the ball.

• Speed:

To find the speed, apply the motion formula:

`v=u+at`

Where:

v is the final velocity = 0 m/s

u is the initial velocity

a is the acceleration due to gravity = -9.8 m/s²

t is the time it takes the ball to reach maximum height = 3/2 = 1.5 seconds.

Rewrite the formula for u and solve:

`\begin{gathered} u=v-at \\ \\ u=0-(-9.8)(1.5) \\ \\ u=14.7\text{ m/s} \end{gathered}`

Therefore, the seed at which he threw the ball is 14.7 m/s.

• Maximum height:

To find the maximum height, apply the formula:

`v^2=u^2+2as`

Where:

s is the maxiumum height.

Thus, we have:

`\begin{gathered} 0^2=14.7^2+2(9.8)s \\ \\ -14.7^2=19.6s \\ \\ s=(-14.7^2)/(19.6)=(216.09)/(19.6) \\ \\ s=11.025\text{ m} \end{gathered}`

Therefore, the ball traveled 11.025 m high.

ANSWER:

• Speed = 14.7 m/s

,

• Maximum height = 11.025 m