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 AB where A and B are two points 20m and 40m below the edge of the tower? ( g=10ms- 2 )

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 AB where A and B are two points 20m and 40m below the edge of the tower? ( g=10ms- 2 ) (A) 2.0 s (B) 1.0 s (C) 0.5 s (D) 0.4 s

Tardigrade Admin · 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 1 0^{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.

Q. 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 AB where A and B are two points 20m and 40m below the edge of the tower? ( g=10ms−2 )

2.0 s

1.0 s

0.5 s

0.4 s

Q. 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? ( $g = 10 m s^{- 2}$ )