Question

A ball is launched vertically, from the ground, with an initial velocity of 25 m/s. When the ball reaches the ground again, what will be its velocity.

Answer 1

Energy conservation principle can be expressed as the sum of the initial kinetic and potential energy in a closed system is equal to the sum of the final potential and kinetic energy in the system

The velocity of the ball when it reaches ground again is 25 m/s

The reason the velocity value is correct is as follows:

The given parameters are;

The direction in which the ball is launched = Vertically upwards

The initial velocity of the ball upwards, v = 25 m/s

Required:

The velocity of the ball when it reaches the ground again

Solution:

By the principle of conservation of energy, we have;

Total mechanical energy, M.E. = Kinetic energy, K.E. + Potential energy, P.E.

M.E. = K.E. + P.E. = Constant

P.E. = m·g·h

When the ball is launched from the ground, h = 0, we have;

P.E. at ground level = 0

K.E. at ground level on the way up, = (1/2)·m·v² = (1/2) × m × 25² = 312.5·m

M.E. = 0 + 312.5·m = 312.5·m = Constant

When the ball is on the way down, at ground level, h = 0, we have

M.E. = 312.5·m = 0 + (1/2)·m·v²

312.5·m = (1/2)·m·v²

v² = 312.5 × 2 = 625

v = √625 = 25

The velocity of the ball when it reaches ground again is v = 25 m/s

Learn more about conservation of energy here:

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