Question #22639

a stone of mass 0.2 kg projected vertically upwards with a velocity of 25 ms calucate the maximum height reached by it above the ground.at what height will it lose 20% of its initial kinetic energy ?

Expert's answer

Kinetic energy at the launch =potential energy at the top

m*v^2 /2 = m*g*h

h = v^2 / (2*g) = (25 m/s)^2 / (2*9.8 m/s^2) = 31.89 m

20% of the initial KE = 0.2* m*v^2 /2

These 20% will be transformed into the potential energy

y = (0.2* m*v^2 /2) / (m*g) = (0.2 * v^2) / (2*g) = (0.2 * (25 m/s)^2 ) / (2 *

9.8 m/s^2) = 6.38 m

Answers: 31.89 m, 6.38 m

m*v^2 /2 = m*g*h

h = v^2 / (2*g) = (25 m/s)^2 / (2*9.8 m/s^2) = 31.89 m

20% of the initial KE = 0.2* m*v^2 /2

These 20% will be transformed into the potential energy

y = (0.2* m*v^2 /2) / (m*g) = (0.2 * v^2) / (2*g) = (0.2 * (25 m/s)^2 ) / (2 *

9.8 m/s^2) = 6.38 m

Answers: 31.89 m, 6.38 m

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