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A stone of mass 0.2 kg is projected upward vertically upwards with a velocity of 25m/s.Calculate the maximum hieght reached by it above the ground. At what height will it lose 20% of its intial K.E?

A stone of mass 0.2 kg is projected upward vertically upwards with a velocity of 25m/s.Calculate the maximum hieght reached by it above the ground. At what height will it lose 20% of its intial K.E?
 

Grade:9

1 Answers

Vikas TU
14149 Points
5 years ago
Maximum height reached = u^2/2g (from projectile formula of max height)
H  = 25^2/2g => 31.88 m
 
Initial K.E = 0.5mv^2 => 0.5*0.2*625 => 62.5 J
20% of 62.5 J  => 12.5 J
From Work energy theorem,
K.E =  U.E
0.5*m*v^2 = mgh
12.5 = 0.2*9.8*h
h = 12.5/(0.2*9.8) = 6.37 m
 
Hence at 6.37 m it would have losen up to 20% of initial K.E.
 
 

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