From the top of a tower of height 50m, a ball is thrown vertically upwards with a certain velocity. It hits the ground 10 s after it is thrown up. How much time does it take to cover a distance A B wh

Question

From the top of a tower of height 50m, a ball is thrown vertically upwards with a certain velocity. It hits the ground 10 s after it is thrown up. How much time does it take to cover a distance A B where A and B are two points 20m and 40m below the edge of the tower? ( g = 10 m s - 2 )

TestHub · Accepted Answer

Let the body be projected upwards with velocity $u$ from top of tower. Taking vertical downward motion of body form top of tower to ground, we have

$u = - u, a = g = 10 m s^{- 2}, s = 50 m, t = 10$ s

As $s = u t + \frac{1}{2} a t^{2}$ ,

So, $50 = - u \times 10 + \frac{1}{2} \times 10 \times 10^{2}$

On solving $u = 45 m s^{- 1}$

If $t_{1}$ and $t_{2}$ are the timings taken by the ball to reach points $A$ and $B$ respectively, then

$20 = 45 t_{1} + \frac{1}{2} \times 10 \times t_{1}^{2}$

and $40 = - 45 t_{2} + \frac{1}{2} \times 10 \times t_{2}^{2}$

On solving, we get $t_{1} = 9.4$ s and $t_{2} = 9.8$ s

Time taken to cover the distance $A B$

$= (t_{2} - t_{1}) = 9.8 = 9.4 = 0.4$ s.

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