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…

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))

MARKS Content Team · Accepted Answer

Correct answer: (D) 0.4 s. 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=10ms^(-2),s=50m,t=10s As s=ut+1/2at^2, So, 50=-u×10+1/2×10×10^2 On solving u=45ms^(-1) If t_1 and t_2 are the timings taken by the ball to reach points A and B respectively, then 20=45t_1+1/2×10×t12 and 40=-45t_2+1/2×10×t22 On solving, we get t_1=9.4 s and t_2=9.8s Time taken to cover the distance AB =t_2-t_1=9.8=9.4=0.4s.

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

Solution

More Motion In One Dimension Questions

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)

Solution

More Motion In One Dimension Questions

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