(i) A stone is thrown vertically upwards with an initial velocity of 60ms. Find a) the maximum height attained by the stone
b) the time taken to reach the maximum height (Take g = 10 ms?)
c) the net displacement if the stone returns back to point from where it was thrown up.

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i) 180m ii) 6sec ii)180m

Explanation:

i) Hmax = U²/2g

= 60²/20

= 3600/20= 180m

ii). Tmax= U/g

= 60/10= 6sec

iii). X= ut²+ 1/2gt²

since U is 0,then ut=0

X=0+ 1/2×10×6²

X=5×36= 180m

hope its clear??

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