A ball thrown vertically upwards with a speed of 19.6 m/s from the top of a tower returns to the earth in 6 seconds. Find the height of tower :
58.8
s=ut+12at2
Let the height of the tower be h.
Here,
u=19.6 m/s
t=6 s
s=−h (The magnitude of the displacement is the height of the tower. Since it is in the direction opposite to the initial velocity, its negative value is taken).
a=−g=−9.8 m/s2 (It is taken as negative because the direction of acceleration due to gravity is downwards, while the initial velocity is upwards).
∴−h=(19.6×6)−9.8 × 6 × 62
=−58.8 m
Thus, the height of the tower is 58.8 m.