JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Kinematics

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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)

Answer: 13

Solution

The ball starts from rest, so after a time

t

its speed is

v = g t

while its acceleration stays

g = 10 m/s^{2}

. The speed equals the numerical value of

g

when

g t = g

, i.e.

t = 1

s. In that time it falls

s = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 1^{2} = 5

m. Measured from the ground, the height at that instant is

18 - 5 = 13

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Integer answer

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