A stone is thrown vertically upwards from the top of a tower 50.0m high with an initial velocity of 20.0ms^(−1) .calculate i.the maximum height the stone reaches ii.the time it takes to reach the maximum height iii.the total distance it covers[take g=10ms^(−1) ]


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Question Number 54015 by shaddie last updated on 28/Jan/19

$A stone is thrown vertically upwards from the top of a tower 50 . 0 m high with an initial velocity of 20 . {0 ms}^{- 1} . calculate$ $i . the maximum height the stone reaches$ $ii . the time it takes to reach the maximum height$ $iii . the total distance it covers [take g = {10 ms}^{- 1}]$
	Answered by estudiante last updated on 29/Jan/19

	$i) a n d i i)$ $l a a l t u r a s e r a l a q u e y a t i e n e m a s l a q u e g a n e . l a a l t u r a f i n a l e s d e l a f o r m a :$ $y = y_{o} + v_{o} t_{a i r e} - \frac{1}{2} {g t}_{a i r e}^{2} = 50 + 20 \times 2 - 0.5 \times 10 \times 4 = 70 m$ $l a v e l o c i d a d f i n a l v e n e l p u n t o m a s a l t o e s c e r o, e l t i e m p o e n e l a i r e t_{a i r e} e s :$ $v = v_{o} - {g t}_{a i r e} \to t_{a i r e} = \frac{v_{o}}{g} = \frac{20}{10} = 2 s$ $i i i)$ $L a d i s t a n c i a t o t a l Y e s :$ $Y = y_{u p} + y_{d o w n} = 20 + 70 = 90 m$
	Commented by estudiante last updated on 28/Jan/19
	May I be right?
	Commented by mr W last updated on 28/Jan/19

	$t o t a l d i s t a n c e c o v e r e d = 20 + 70 = 90 m$
	Commented by estudiante last updated on 29/Jan/19
	Ooops :) You're right. I will change it. Thanks
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