Question

A ball is projected from the bottom of a tower and is found to go above the tower and is caused by the thrower at the bottom of the tower after a time interval t1. An observer at the top of the the tower sees the same ball go up above him and then come back at this level in a time interval t2 the height of the tower is

Answer 1

Hey Dear,

◆ Answer -

h = 5/2 t1(t1-t2) - 5/4 (t1-t2)²

● Explanation -

Time taken by ball to reach max height from bottom is t1/2.

From bottom of tower to max height,

v = u + at

0 = u + (-10)×t1/2

u = 5t1

Similarly time taken by ball to reach max height from top is t2/2.

Thus, time taken by ball to reach top from bottom of tower will be

t' = t1/2 - t2/2

t' = (t1-t2)/2

Now the height of tower h is calculated as -

h = ut' + 1/2 at'²

h = 5t1 × (t1-t2)/2 + 1/2 × (-10) × [(t1-t2)/2]²

h = 5/2 t1(t1-t2) - 5/4 (t1-t2)²

Hope this helps you...

Answer 2

Explanation:

this is the correct answer