Question

A body released from the top of a tower takes 6 seconds to reach the ground. After time of 3seconds from starting, its height from the ground is ( g=10m/s2)​

Answer 1

Answer:

After 3  sec position from the ground = 135m

Explanation:

The body was released from the top of a Tower hence initial velocity was ZERO and gravitational acceleration = $g = 10m/s^2$

Height = Distance covered in 6 seconds

$S=ut+\frac{1}{2}gt^2, u=0, g=10m/s^2, t=6sec\\\\S=0*6+\frac{1}{2}*10*6^2 = 180m$

Height = 180 m

Now in 3 seconds it will reach from top of the tower

$S=ut+\frac{1}{2}gt^2, u=0, g=10m/s^2, t=3sec\\\\S=0*3+\frac{1}{2}*10*3^2 = 45m$

Therefore its position from the ground = 180m - 45m = 135m