A man on top of a tower of height 35m throws a stone vertically upwards with a speed of 14m/s. Find: (i)the height above the ground, reached by the stone. (ii)the speed of the stone,when it reaches the ground.


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Question Number 19890 by NECC last updated on 17/Aug/17

$A m a n o n t o p o f a t o w e r o f h e i g h t$ $35 m t h r o w s a s t o n e v e r t i c a l l y$ $u p w a r d s w i t h a s p e e d o f 14 m / s .$ $F i n d :$ $(i) t h e h e i g h t a b o v e t h e g r o u n d,$ $r e a c h e d b y t h e s t o n e .$ $(i i) t h e s p e e d o f t h e s t o n e, w h e n$ $i t r e a c h e s t h e g r o u n d .$
	Answered by mrW1 last updated on 17/Aug/17

	$(i)$ $mg Δ h = \frac{1}{2} {mv}_{0}^{2}$ $Δ h = \frac{v_{0}^{2}}{2 g} = \frac{14^{2}}{2 \times 9.81} \approx 10 m$ $h = h_{T} + Δ h = 35 + 10 = 45 m$ $(ii)$ $mgh = \frac{1}{2} {mv}^{2}$ $v = \sqrt{2 gh} = \sqrt{2 \times 9.81 \times 45} \approx 29.7 m / s$
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