Question

A ball is dropped from the roof of a tower of height h. The total distance covered by it in the last second of its motion is equal to the distance covered by it in first three seconds, what is the value of h? (g = 10 m/s2

Answer 1

Answer:

h = ut +0.5 gt2

as u =0 and g =10 m/s2

Let total height = H

Total time = T

H = 0 5 gT2 = 5 T2

Distance travelled in nth second ( last second )

Sn = 0.5 g ( n2 — ( n-1)2 ) = 0.5 g ( 2n-1) = H/2

So 0.5 gT2 /2 = 0.5 g ( 2n -1) = 5 (2n -1 )

H/2 = 5 (2n-1)

Put the value of H equal to 5 T2 and n = T

T2 /2 = ( 2T-1)

T2 = 2 ( 2T-1)

T2 = 4T -2

T2 -4T +2 = 0

Using quadratic solutions

T can be 3.415 Or 0.585

Now 0.585 is not possible

So T = 3.415 second