Assume the force of gravity to be constant for short heights above the surface of the Earth. A body is dropped from rest at a height h. If we ignore air resistance, what will its kinetic energy be just before it hits the ground?

\sqrt{2 h}

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\sqrt{g h}

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\sqrt{2 g h}

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\sqrt{4 g h}

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Solution

The correct option is C $\sqrt{2 g h}$
W = F · ∆r = = mgh
$Δ K = K_{f} - K_{i} = [\frac{1}{2} m v_{f}^{2}] - 0 = \frac{1}{2} m v_{f}^{2}$
W = ∆K
mgh = $\frac{1}{2} m v_{f}^{2}$
v = $\sqrt{2 g h}$

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Assume the force of gravity to be constant for short heights above the surface of the Earth. A body is dropped from rest at a height h. If we ignore air resistance, what will its kinetic energy be just before it hits the ground?

The correct option is C √2ghW = F · ∆r = = mgh ΔK=Kf−Ki=[12mv2f]−0=12mv2f W = ∆K mgh = 12mv2f v = √2gh

The correct option is C $\sqrt{2 g h}$
W = F · ∆r = = mgh
$Δ K = K_{f} - K_{i} = [\frac{1}{2} m v_{f}^{2}] - 0 = \frac{1}{2} m v_{f}^{2}$
W = ∆K
mgh = $\frac{1}{2} m v_{f}^{2}$
v = $\sqrt{2 g h}$