Question

A ball is thrown up to a top height of 12.0m above a person's hand, what was the initial velocity of the ball?

Answer 1

Given that the ball is thrown to a top height of 12.0 meters above a person's hand.

Let's find the initial velocity of the ball.

To find the initial velocity, apply the motion formula:

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

Where:

• s is the distance = 12.0 m

,

• v is the final velocity.

Here, the final velocity is 0 m/s.

• u is the initial velocity.

a is the opposite of acceleration due to gravity = -9.8 m/s^2

Let's solve for u.

Hence, we have:

`\begin{gathered} 0^2=u^2+2(-9.8)(12) \\ \\ 0^2=u^2-235.2 \\ \\ u^2=235.2 \\ \\ \text{Take the square root of both sides:} \\ u=\sqrt[]{235.2} \\ \\ u=15.3\text{ m/s} \end{gathered}`

Therefore, the initial velocity is 15.3 m/s

ANSWER:

15.3 m/s