Question

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) the ball’s initial velocity and (b) the height it reaches.

Answer 1

Answer:

Initial velocity = $29.43m/s$ b) Height it reaches = 44.145 m

Explanation:

Using the first equation of motion we have

$v=u+at$

here

v is the final velocity

u is the initial velocity

a is the acceleration of the object

t is time

When the ball reaches it's highest point it's velocity will become 0 as it will travel no further

Also the acceleration in our case is acceleration due to gravity ($-9.8m/s^{2}$) as the ball moves in it's influence alone with '-' indicating downward direction

Thus applying the values we get

$0=u-(9.81)m/s^{2}\times 3\\\\u=19.43m/s$

b)

By 3rd equation of motion we have

$v^{2}-u^{2}=2gs$

here s is the distance covered

Applying the value of u that we calculated we get

$s=\frac{-u^{2}}{2g}\\\\s=\frac{-29.43^{2}}{-2\times 9.81}\\\\s=44.145m$