Question

a body released from the top of a tower falls through half the height of the tower in 3 seconds it will reach the ground after nearly seconds​

Answer 1

A body released from the top of a tower will reach the ground after 4.24 seconds

1. Using kinematics equation calculating the height of the tower
2. $S= ut + \frac{1}{2}at^{2}$
3. Here S is the distance travelled = h/2 , u is initial velocity =0,a is the acceleration = acceleration of gravity (g =9.8 m/s) and t is the time taken to cover h/2 distance = 3 s
4. $\frac{h}{2} = 0 + \frac{1}{2}9.8(3^{2})$
5. Height of the tower h= 9.8 x 9 =88.2 meter
6. Finding the time taken to reach the ground from top of the tower (T)
7. $h = uT + \frac{1}{2}g(T^{2})$
8. $88.2 = 0.T + \frac{1}{2}9.8(T^{2})$
9. $T= \sqrt{\frac{88.2 X 2}{9.8} } = 4.24 s$