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):

Question

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): (A) 120 m (B) 135 m (C) 175 m (D) 180 m

Tardigrade Admin · Accepted Answer

Let the ball reaches top of tower of height h at times t1 and t2 then t2- t1 = 6 and t1 + t2 = 12. Solving them: t1 = 3 s, t2 = 9 s Now height of tower: h = (1/2) gt1t2 = (1/2) × 10 × 3 × 9 = 135 m

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

120 m

135 m

175 m

180 m

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:

$h = \frac{1}{2} g t_{1} t_{2}$

$= \frac{1}{2} \times 10 \times 3 \times 9$

$= 135 m$

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 6s and the ball reaches the ground after 12s. The height of the tower is (g=10m/s2):