Question

A stone is dropped from the top of a 20m.tower.What will be its velocity just before touching the surface of the earth ? Calculate time taken by the stone to reach the earth's surface.Take the value of "g" as 10m/s2.

Answer 1

{1}.height of tower h=20m.
2.initial velocity of the stone u = 0
g=10m/s2
s=ut +1/2gt2
=>20=0×t+1/2 ×10t2
=>5t2=20
=>t2= 4
=>t=√4
=>t=2s
Hence,time taken by the stone to reach the earth's surface is 2sec.
{2.}Velocity of the stone :-
v = u +gt
=u + 10×2m/s (u=0)
=20m/s
Hence,the speed of the stone reaching the earth's surface is 20ms-1.

Answer 2

for the time
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s  = ut + 1/2 gt²
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s = 20 m
u = 0
g = 9.8
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20  = 0 + 1/2 (10)t²

20 = 5 t²

t² = 4.
t = 2.
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therefore time taken = 2 seconds
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for the final velocity
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v = u + gt
---
u = 0
g =9.8
t =2
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v = 0 + 10(2)

v = 20 m/s
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its velocity just before touching the surface of the earth = 20 m/s