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Grade: 12
        
Que. A stone is thrown vertically upward with velocity of 20 m per second from 25 m top of Tower. Determine - (1) the distance covered by stone before reaching the ground. (2) maximum height attained during flight. (3) final velocity of stone before reaching the ground.
2 years ago

Answers : (1)

aditya kulkarni
53 Points
							
the stone is thrown with initial velocity 20 m/s
it is thrown under gravity so gravitational accleration will be9.8 m/s and final velocity will be 0
therefore bu using v2=u2+2as where v=final velocity, u=initial velocity, a= gravitational acceleration, s= distance covered.
after putting values in the eqn and solving we get distance covered=s=20.4 m
so the stone has covered a dist of 20.4m after throwing in the vertical direction
so its new height is 25+20.4=45.4m this the max height
before reaching the ground it will travel a distance of 20.4+20.4+25=65.8m
now for its downward journey, 
initial velocity=u=0
final velocity=v
acceleration=9.8 
dist to be covered = 45.4m
therefore putting above values in the eqn     v2=u+2as
on solving we get   v2=889.84
so v=29.83
2 years ago
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