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The ball is thrown vertically upwards with a velocity u. Let the mass of the ball be m. The kinetic velocity of the ball is (1/2)*m*u^2.
As it rises, the kinetic energy is converted to potential energy given by m*g*h. At the highest point, all the kinetic energy has been converted to potential energy.
Here, we have m*g*h = (1/2)m*u^2
=> h = u^2/2g
When the ball starts to drop down, the direction of its velocity is opposite that with which it rose up. By the time it reaches the bottom, all the potential energy it had acquired at the height u^2/2g has been converted to kinetic energy. If its velocity is v, we have
m*g*(u^2/2g) = (1/2)*m*v^2
=> u^2 = v^2
=> v = -u ( as they have an opposite direction)
The velocity with which it falls back to the Earth is -u.
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