Question

A hiker stands on the edge of a cliff 490 m above the ground and throw a stone horizontally with a speed of 15 m s^-1. The speed with which the stone hits the ground is

• A. 15 m s^-1
• B. 90 m s^-1
• C. 99 m s^-1
• D. 49 m s^-1

Answer 1

Answer: (C)

99 m s^-1

Answer 2

Motion along horizontal directions.

$\downarrow + \mathrm { ve }$

$u_{x} = 15ms^{- 1},a_{x} = 0$

$v_{x} = u_{x} + a_{x}t = 15 + 0 \times 10 = 15ms^{- 1}$

Motion along vertical directions,

$u_{y} = 0,a_{y} = g$

$v_{y} = u_{x} + a_{x}t = 0 + 9.8 \times 10 = 98ms^{- 1}$

$\therefore$Speed of the stone when it hits the ground is

$v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{(15)^{2}(98)^{2}} = 99ms^{- 1}$