Kinematics

JEE Main level, numerical answer, medium

152 of 308

Question 152 of 308

From 1818 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=10g = 10 m/s2^2 and neglect the air resistance)

Show answer and solution

Answer: 13

Let the ball fall a distance ss from the initial height.

Initial velocity u=0u = 0

Acceleration a=g=10a = g = 10 m/s2^2

Given that the magnitude of velocity is equal to the magnitude of acceleration due to gravity, we have v=10v = 10 m/s.

Using the equation of motion v2=u2+2gsv^2 = u^2 + 2gs:

(10)2=02+2×10×s(10)^2 = 0^2 + 2 \times 10 \times s

100=20s100 = 20s

s=5s = 5 m

The distance fallen by the ball is 55 m.

The height above the ground is h=H−sh = H - s

h=18−5=13h = 18 - 5 = 13 m

Answer: 1313

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