Question

A baseball is popped straight up into the air and has a hang-time of $6.25s$. Determine the height to which the ball rises before it reaches its peak.

• A. $48.0m$
  
• B. $58.0m$
  
• C. $78.0m$
  
• D. $88.0m$
  

Answer 1

Answer: (A)

$48.0m$

As the hang time is the total time to reach peak and come back, so the time for reaching to peak is half of the hang time i.e. $t=6.25/2=3.125s$

When the ball reaches to peak point, the final velocity will be zero. i.e. $v=0m/s$

Here, the acceleration of ball is $a=-g=-9.8m/s^2$ (minus for motion against gravity)

let $u$ be the initial velocity of the baseball.

Using $v=u+at$,

$0=u+(-9.8)\left(3.125\right)$

$⟹$ $u=30.625m/s$

If $h$ be the height reached by ball.

Using $v^2-u^2=2ah$

$0^2-(30.625{)}^2=2(-9.8)h$

$⟹$ $h=47.85m\sim 48m$