Particle A is projected vertically upward from the ground. Exactly 2 s later, particle B is released from rest from a height of 50 m. Taking g=10 m s⁻², find the minimum projection speed of A so that the two particles meet while A is still moving upward, in m s⁻¹.
Physics · Kinematics Part-1 (one dimension) · Motion under gravity
JEE Mains JEE Advanced

Question

Particle A is projected vertically upward from the ground. Exactly 2 s later, particle B is released from rest from a height of 50 m. Taking g=10 m s⁻², find the minimum projection speed of A so that the two particles meet while A is still moving upward, in m s⁻¹.

Physics · Kinematics Part-1 (one dimension) · Motion under gravity

G1 Physics · Accepted Answer

Use the limiting case in which A meets B exactly at the highest point of A. With τ=2 s, H=50 m, g=10 m s⁻², minimum u = [gτ+√(4gH−g²τ²)]/2 = [20+√(2000−400)]/2 = [20+40]/2 = 30 m s⁻¹.