# a ball is thrown vertically upward with a speed of 19.6 m/s from the top of a tower and returns to earth in 6 sec find the height of the tower

@geethanjali : Correct! Keep working towards your studies and you will excel in your exams! u = 19.5 m/s

t = 5 sec

g = 9.8 m/s²

Total distance covered while going up = S

total distance covered while coming down = h+S

At topmost point

v = 0

v = u - gt

0 = 19.5 - 10 (9.8)

t = 2 sec (appx)

Now,

The velocity of ball when it falls down a distance S after reaching topmost height (h+S) will be same as initial velocity (calculate yourself), and the time taken will again be 2 sec.

So, total time elapsed = time for going up S + time for coming down S = 2+2 = 4 sec

Total time = 5 sec

Remaining time = 5-4 = 1 sec

here u = 19.5 m/s(downwards) , t = 1 sec

h = ut + 1/2 g t²

h = 19.5 (1) + 1/2 (9.8) (1) = 19.5 + 4.9 = 24.4 m

• 3

Takng vertical downwards motion of the ball frm point of projection up to ground, we have

u = -19.6 m/ sec

a = 9.8 m / sec square

t = 6 sec..

s = ut + 1/2 at square

s = - 19.6 * 6 + 1/2 * 9.8 * 6 * 6

s = 58.8 m ans..

HOPE U GET IT...

• 79
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