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:
[tex]v^2=u^2+2as[/tex]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:
[tex]\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}[/tex]Therefore, the initial velocity is 15.3 m/s
ANSWER:
15.3 m/s
