A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the bottom of the cliff? (b) What is its speed just before hitting? And (c) what total distance did it travel? (Ans: 5.2 sec, 38.94 m/s, 84.7 m)

Answer

given,

vertical speed of stone, v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

v = u + a t

v = 12 - 9.8 x 5.2

v = - 38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

vertical distance covered by the stone

v² = u² + 2 g h

0 = 12² - 2 x 9.8 x h

h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.