A particle is thrown vertically upwards with velocity 11.2 kms⁻¹ from the surface of earth. Calculate its velocity at height 3R, where R is the radius of earth.

The given velocity equals escape velocity. So, GMm/R = initial K.E. v_escape = √(2GM/R) Initial total energy = K.E + P.E = -GMm/R + GMm/R = 0 So, at height 3R which is equal to distance 4R from the centre of the earth, total energy will be zero. Hence, 0 = -GMm/4R + 1/2mv² ⇒ v = √(GM/2R) = v_escape/√2 On solving, we get v = 5.6 m/s