Question

A ball is dropped from certain height such that the
distance travelled in last second is equal to the
distance travelled in first three second. Find its
height
del​

Answer 1

Answer:

The height will be 125 m

Explanation:

Let the The distance covered in first 3 second  as d,

then,

d=1/2gt^2

=1/2×10×3^2

=45

If the ball takes 'p' second to fall to ground . Let the distance covered in pth second  d(p)

then,

d(p)=u+g/2(2p−1)

=0+10/2(2−1)

=10p-5

According to the problem distance covered in last second is equal to the the distance in first three second

Therefore 45=10p−5

p=5

ley the height as h,

Therefore h=1/2gt^2

=1/2×10×25

=125m