Question

A ball is thrown straight up into the air. A stop watch measures that time the ball stays in the air until it comes back down to the point from which it was thrown is 6s.
a. How high does the ball go?
b.with what speed was the ball thrown initially? What speed does it return to its initial height?

Answer 1

HINTS: A ball takes the same amount of time to go up and come back down! :D
Final Velocity at highest point is 0 [While thrown]
Initial velocity at highest point is 0[When it is coming down]
Acceleration while going up is -9.8m/s2
Acceleration due to gravity is 9.8m/s2

Now can you calculate? :D

Answer 2

@PaaSolo ?

Answer 3

so which formula do i use

Answer 4

WHEN IT IS GOING UP

First use v=u + at to find the initial velocity

Then use

s = ut + at^2/2

To get the distance traveled upwards. :)

Understood? :)

Answer 5

ok thanks

Answer 6

:)

Answer 7

hi sorry for botgering yoyu but what will be the t and t^2 in the second equation for part a.