RRB Group D 1 Dec 2025 Shift 1 Solved Paper

Concept:Use the third equation of motion for vertical motion under constant gravity.Explanation:At maximum height, the object momentarily stops, so final velocity $v = 0$.The relevant kinematic equation is $v^2 = u^2 - 2gh$, where $u$ is the initial speed, $g$ is acceleration due to gravity, and $h$ is the maximum height.Substitute the given values: $v = 0$, $g = 10 \, \text{m/s}^2$, $h = 20 \, \text{m}$.This gives $0 = u^2 - 2 \times 10 \times 20$.Simplify: $0 = u^2 - 400$.Thus $u^2 = 400$, and $u = \sqrt{400} = 20 \, \text{m/s}$.Therefore, the initial speed of the object is $20 \, \text{m/s}$.Answer:20 m/sec