Question

The particle is dropped from the top of the tower the distance covered by it in the last one second is equal to that cover by heat in the first 3 seconds find the height of the tower

Answer 1

Answer:

The height of the tower is 125 meter.

Explanation:

let g= 10 m/s^2

h= height of the tower.

Let, the distance covered in first 3 second  is d, then

d=1/2gt^2

=1/2×10×3^2 [where g= 10m/s^2 , t= 3 s]

=45

Now,the particle is dropped from the top of the tower the distance covered by it in the last one second is equal to that cover by heat in the first 3 seconds.

Again, let the ball takes p second time to reach to ground .

The distance covered in pth second

d(p)=u+g/2(2p−1)

=0+10/2(2p−1)

=10p−5

Therefore

45=10p−5

or,p=5

Therefore h=1/2gt^2

=12×10×5^2

=125m