Question

8. A hiker stands on the edge of a cliff 490 m above the
ground and throws a stone horizontally with an intial speed
of 15 ms-1. The speed with which it hits the ground is
(a) 99 ms
(b) 101 ms
(c) 103 ms
(d) 105 ms
-1
- 1​

Answer 1

motion of stone may be considered as the super position of the two independent sources given horizontal with constant velocity u=15 m/s

vertices motion with constant acceleration a=g=9.8 m/s2

Let h be the height of the cliff above ground

Let uv be the vertical component of velocity of projection of the stone If the stone hits the ground after 1 second of projection the h=uvt+21gt2

the stone is thrown horizontally vertical component 

∴h=0+21gt2⇒t=92h

This gives time for stone to reach ground

t==2h99.82×490=9.8980=100=10Δ

t=10Δ

Let V4 be the vertical component of velocity of stone when it hots the ground then

vy2±uy2+2gh

vy2=(0)2+2×+9.8×490

vy