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
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