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 )

Step-by-step solution

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 + 1 2 a t 2 , So, 50 = - u × 10 + 1 2 × 10 × 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 + 1 2 × 10 × t 1 2 and 40 = - 45 t 2 + 1 2 × 10 × t 2 2 On solving, we get t 1 = 9.4 s and t 2 = 9.8 s Time taken to cover the dista