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-th

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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