Question

A hiker stands on the edge of a cliff $490\,m$ above the ground and throws a stone horizontally with a speed of $15\,ms^{-1}$, the speed with which the stone hits the ground is

• A. $15\,ms^{-1}$
• B. $90\,ms^{-1}$
• C. $99\,ms^{-1}$
• D. $49\,ms^{-1}$

Answer 1

Answer: (C)

Motion along horizontal direction,
$u_x=15\,ms^{-1}$, $a_x=0$
$v_x = u_x + a_xt$
$= 15 + 0 \times 10$
$ = 15\,ms^{-1}$
Motion along vertical direction,
$u_y=0$, $a_y=g$
$v_y = u_x + a_yt$
$= 0+9.8 \times 10$
$ = 98\,ms^{-1}$
$\therefore $ Speed of the stone when it hits the ground is
$v=\sqrt{v^{2}_{x}+v^{2}_{y}}$
$=\sqrt{\left(15\right)^{2}+\left(98\right)^{2}}$
$\approx99\,ms^{-1}$