Question

a stone
dopped from the top of a building reaches the ground in 0.5 sec.
calculate
(i)before it reaches the ground.
(ii) height of the building from the ground. (g=10m/sec2)
​

Answer 1

1:- the velocity just before reaching the ground is 5m/s

2:- height of building is 1.25 m.

Answer 2

u=initial speed = 0.

t = time = 0.5 sec.

a= acceleration = g = 9.8 m/sec^2

findout distance:-

s=ut+ \frac{1}{2} at^{2}

s=0*t + \frac{1}{2}*9.8* \frac{1}{2}* \frac{1}{2}= \frac{9.8}{8}=1.225m

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