A ball is dropped from rest from a height of 18 m above the ground. Taking g = 10 m·s⁻² and neglecting air resistance, at what height above the ground is the numerical magnitude of the velocity equal to the numerical magnitude of the acceleration? Give the answer in metres.
Physics · Kinematics part-2 Projectile Motion · Projectile from height
JEE Mains 2026 5 April Morning Shift

Question

A ball is dropped from rest from a height of 18 m above the ground. Taking g = 10 m·s⁻² and neglecting air resistance, at what height above the ground is the numerical magnitude of the velocity equal to the numerical magnitude of the acceleration? Give the answer in metres.

Physics · Kinematics part-2 Projectile Motion · Projectile from height

JEE Mains 2026 5 April Morning Shift

G1 Physics · Accepted Answer

The acceleration magnitude remains constant: a = g = 10 m·s⁻². The velocity after time t is v = gt. For the numerical values of v and a to be equal, 10t = 10, so t = 1 s. The distance fallen in 1 s is s = (1 ÷ 2)gt² = (1 ÷ 2)(10)(1)² = 5 m. Therefore the height above the ground is 18-5 = 13 m. Hence the answer is 13.