Question

From the top of a tower of height 50 m, 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 20 m and 40 m below the edge of the tower? (g = 10 m/s2)
1) 2.0 s
2) 1.0 s
3) 0.5 s
4)0.45​

Answer 1

Answer:

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Answer 2

Answer:

Time taken to cover the distance AB = 0.4sec

Explanation:

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

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

as,     s = ut + 1/2a$t^{2}$

so,    50 = -u × 10+ 1/2 × 10  ×$10^{2}$

On solving, u = 45m/s

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

      20 = 45$t_{1}$ + 1/2 × 10 × $t_{1}^{2}$

and 40 = -45$t_{2}$ + 1/2 × 10 × $t_{2} ^{2}$

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

Time taken to cover the distance AB = ($t_{2}$ - $t_{1}$)

                                                              = 9.8 - 9.4

                                                              = 0.4s