A hiker stands on the edge of a cliff $490 ext{ m}$ above the ground and throws a stone horizontall

Question

A hiker stands on the edge of a cliff $490	ext{ m}$ above the ground and throws a stone horizontally with an initial speed of $15	ext{ m/s}$. Neglecting air resistance, what is the speed with which it hits the ground? (Take $g = 9.8	ext{ m/s}^2$)

Professor PC Thomas Classes · Answer

Time of flight $t = \sqrt{2h/g} = \sqrt{2(490)/9.8} = 10	ext{ s}$. Final velocity components: $v_x = 15	ext{ m/s}$, $v_y = gt = 9.8 	imes 10 = 98	ext{ m/s}$. Final speed is $\sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 98^2} = \sqrt{225 + 9604} \approx 99	ext{ m/s}$.