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 $(g = 10 m / s^{2})$

120 m

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135 m

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175 m

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80 m

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Solution

The correct option is B $135 m$

According to question,
$t_{B C} = \frac{6}{2} = 3 s$
$t_{A C} = \frac{12}{2} = 6 s$
$∴ t_{A B} = 3 s$
By using first equation of motion; as at the top $(v = 0)$
$∴ 0 = u - (10) 6$
or $u = 60 m / s$
When we throw the ball upward then the distance travelled by ball in initial 3 second is equal to the height to tower
i.e., $h = u t_{A B} - \frac{1}{2} g t_{A B}^{2}$
$= (60) (3) - \frac{1}{2} (10) (3)^{2}$
$= 135 m$

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