A ball is thrown vertically upward from the ground level with velocity of 5 m/s. The final velocity of the ball just before it hits the ground is the same with its initial velocity 5 m/s.
Using equation of motion:
v² = u² - 2gh
Where:
v = final velocity
u = initial velocity
h = maximum height
g = acceleration due to gravity
Since the ball stops at the maximum height, v = 0. Hence,
u² = 2gh
h = u² / 2g = 5² / 2g = 12.5 / g
Using the law of conservation of energy:
Ek1 + Ep1 = Ek2 + Ep2
Let:
1 = position at the maximum height
2 = position before the ball hits the ground
Hence,
0 + mgh = 1/2 mv² + 0
v² = 2gh = 2 x g. 12.5/g
v² = 25
v = 5 m/s
Hence, the final velocity of the ball just before it hits the ground is the same with initial velocity it is thrown.
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