Question

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

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

Answer 1

Answer: (C)

Motion along horizontal direction,
$u_x=15m{s}^{-1}$, $a_x=0$
$v_x=u_x+a_{x}t$
$=15+0\times 10$
$=15m{s}^{-1}$
Motion along vertical direction,
$u_y=0$, $a_y=g$
$v_y=u_x+a_{y}t$
$=0+9.8\times 10$
$=98m{s}^{-1}$
$\therefore$ Speed of the stone when it hits the ground is
$v=\sqrt{v_x^2+v_y^2}$
$=\sqrt{{\left(15\right)}^2+{\left(98\right)}^2}$
$\approx 99m{s}^{-1}$