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  4. A ball is projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of 6 s and the ball reaches the ground after 12 s. The height of the tower is (g = 10 m/s2):

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Q. A ball is projected upwards from the foot of a tower. The ball crosses the top of the tower twice after an interval of $6 \,s$ and the ball reaches the ground after $12 \,s$. The height of the tower is $(g = 10 \,m/s^2)$:

Motion in a Straight Line

Solution:

Let the ball reaches top of tower of height $h$ at times
$t_1$ and $t_2$ then $t_2- t_1 = 6$ and $t_1 + t_2 = 12$.
Solving them: $t_1 = 3 \,s, t_2 = 9 \,s$
Now height of tower:
$h = \frac{1}{2} gt_1t_2$
$ = \frac{1}{2} \times 10 \times 3 \times 9$
$ = 135 \,m$