Motion Under Gravity

Question

A stone is dropped from rest at the top of a tower and reaches the ground after 3 seconds of free fall. Taking g = 10 m/s^2 and neglecting air resistance, what is the height of the tower?

RankGuru · Accepted Answer

Free fall is a special case of uniformly accelerated motion in which the only acceleration is that due to gravity, directed downward with magnitude $g$. The relevant equation of motion is $h = ut + 	frac{1}{2}gt^2$, where $h$ is the vertical distance fallen. Since the stone is dropped, its initial velocity $u = 0$, so the relation reduces to $h = 	frac{1}{2}gt^2$. Substituting $g = 10	ext{ m/s}^2$ and $t = 3	ext{ s}$ gives $h = 	frac{1}{2}(10)(3^2) = 	frac{1}{2}(10)(9) = 45	ext{ m}$. The option 30 m is wrong because it mistakenly uses $gt$ rather than the quadratic term. The option 90 m omits the factor of one-half. The option 15 m incorrectly takes $	frac{1}{2}gt$ without squaring the time. This is the classic NCERT free-fall result for a body released from rest. As a sanity check, the distance grows with the square of time, so doubling the fall time would quadruple the height, confirming the strong time dependence characteristic of constant-acceleration descent.

Motion Under Gravity

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