JEE Main 2026 — Apr 5, Shift 1 — Physics solution

JEE Main2026Apr 5, Shift 1Physics

From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ___________ m.

(Take $g = 10$ m/s $^{2}$ and neglect the air resistance)

Your answer

Solution

From rest, the speed at time $t$ is $v = g t$ , while the acceleration stays constant at $g$ . Setting magnitudes equal (numerically, in the same units): $g t = g \Rightarrow t = 1$ s.

Distance fallen in that second: $\frac{1}{2} g t^{2} = \frac{1}{2} (10) (1)^{2} = 5$ m.

Height above ground: $18 - 5 = 13$ m.

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