A ball thrown vertically upwards with a speed of 20m/s from the top of a tower reaches a certain height and then falls back to the earth in 6 s. Find the height of the tower. [Take acceleration due to gravity, g=10 m/s2]
160 m
Let the height reached by the ball when thrown upwards be H and the height of the tower be h.
Here,
Initial velocity, u=20 m/s
Final velocity after reaching height, H is v=0
acceleration, a=−g=−10 m/s2 [It is taken as negative because the direction of acceleration due to gravity is downwards, while the initial velocity is upwards]
Using, v2=u2+2as
0=202−2gH
H=2022g=20×2020=20 m
Now, the ball will fall down from a height of 20 m from above the tower.
Initial velocity, u′=0
time, t=6 s
displacement, s=20+h
acceleration, a=g=10 m/s2
Using s=ut+12at2,
20+h=0×6+12×10×6×6
⇒20+h=180
⇒h=180−20=160 m
Thus, the height of the tower is 160 m