Question

A ball is thrown upwards at an unknown speed. in a time of

.5 second the ball has a height of 6 meters.

what is the initial speed of the ball?

what is the balls speed at a height of 6 meters?

what would be the balls maximum height? (hint: you dont know the time to reach maximum height)

Answer 1

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Step-by-step explanation:

a) We have equation of motion
`s=ut+(1)/(2) at^2`, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8
`m/s^2`

Substituting

`6=u*0.5-(1)/(2) *9.8*0.5^2\\ \\ u=14.45m/s`

Initial speed of the ball = 14.45 m/s

b) We have equation of motion
`v^2=u^2+2as`, where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8
`m/s^2`

Substituting

`v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s`

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion
`v^2=u^2+2as`, where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8
`m/s^2`

Substituting

`0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m`

Maximum height reached = 10.65 m