A particle is projected vertically upward from the ground with speed 45 m s⁻¹. Taking g=10 m s⁻², find the actual distance travelled during the 5th second, in metre.
Physics · Kinematics Part-1 (one dimension) · Motion under gravity
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Question

A particle is projected vertically upward from the ground with speed 45 m s⁻¹. Taking g=10 m s⁻², find the actual distance travelled during the 5th second, in metre.

G1 Physics · Accepted Answer

The 5th second is the interval 4 s to 5 s. The particle reaches its highest point at t = u/g = 45/10 = 4.5 s. So the turning point lies inside the 5th second. From 4 s to 4.5 s: velocity changes from 5 m s⁻¹ to 0. Distance = (5+0)(0.5)/2 = 1.25 m. From 4.5 s to 5 s: speed increases from 0 to 5 m s⁻¹ downward. Distance = (0+5)(0.5)/2 = 1.25 m. Total distance = 2.5 m.