Question

A ball is dropped from certain height. if it takes 0.2 s to cross the last 6.2 m before hitting the ground, find the height from which it was dropped. (take g=10 ms^-2)

Answer 1

See the figure:

let the total height be x

and, first height (h1) = (x - 6.2)

given, last height (h2) = 6.2m

t2 ( t taken to cover 6.2m)= 0.2s

u1 = 0 m/s

a = (g) = (10 m/s^2)

Now, we will find u2

s = ut + 1/2 × gt^2

=> 6.2 = u × 0.2 + 1/2 × 10 × (0.2)^2

=> 6.2 = 0.2u + 5 × 0.04

=> 6.2 = 0.2u + 0.2

=> 6 = 0.2u

=> 30 = u

[u2 = 30 m/s]

u2 = v1 (Since , we have assumed both on the same point)

v1 = 30 m/s

we will find the time taken to cover h2;

v = u + at

=> 30 = 0 + 10t

=> 3 = t

[t = 3s]

Now ,

s = ut + 1/2 × gt^2

=> x - 6.2 = 0 × 3 + 1/2 × 10 × (3)^2

=> x - 6.2 = 5 × 9

=> x = 51.2 m