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

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

Step-by-step 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)* 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=(-u^(2))/(2g)\\\\s=(-29.43^(2))/(-2* 9.81)\\\\s=44.145m`