Question

You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the ball reaches relative to the throwing point How long does it take the ball to reach the height What is the position of the ball at t=2s? At what height does the ball have a speed of +5m/s?

Answer 1

Step-by-step explanation:

Given

initial velocity
`u=11.2\ m/s`

At maximum height velocity of ball is zero

using

`v^2-u^2=2as`

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

`(0)^2 -(11.2)^2=2* (-9.8)* (s)`

`s=(11.2^2)/(2* 9.8)`

`s=6.4`

time taken by the ball to reach the maximum height

`v=u+at`

`0=11.2-9.8* t`

`t=(11.2)/(9.8)`

`t=1.142\ s`

At
`t=2\ s` height of ball is

`h=ut+\\frac{1}{2}at^2`

`h=11.2* 2-(1)/(2)9.8* (2)^2`

`h=22.4-19.6`

`h=2.8\ m`

i.e. ball is moving downward

height at
`v=5\ m/s`

`v^2-u^2=2as`

`s=(25-125.4)/(2* (-9.8))`

`s=5.12\ m`