Question

A ball is dropped from height h.The total distance covered by the body in last second of its motion is equal to distance covered by it in the first three seconds ,what is the vslue of h.

Answer 1

distance covered in first 3 seconds

using equation

S=ut+a(t)2/2

u=0.(given)

S=10(3)2/2=45m

let last second be nth second

distance covered in nth sec =distance covered in n seconds-distance covered in (n-1) seconds

Sn- Sn-1=
1/2an^2 -1/2a(n-1)^2=1/2a*(2n-1)

so,

a(2n-1)/2=45

10(2n-1)/2=45

2n-1=9

n=5

to find the height h

we have to find the distance covered in 5 seconds

S5=ut+a(t)2/2

S5=(0)5+10(5)2/2

S5=125m=h