Collision Velocity Squared – Rankers Physics
Topic: Kinematics
Subtopic: Relative Motion in One Dimension

Collision Velocity Squared

A body A is thrown up vertically from the ground with velocity \(v_0\) and another body B is simultaneously dropped from a height H. They meet at a height \(H/2\) if \(v_0^2\) is equal to
\(2gH\)
\(gH\)
\((1/4)gH\)
\((2g)/H\)

Solution:

Position of B: \(y_B(t) = H - (1/2)gt^2\). Meeting at \(H/2\): \(H/2 = H - (1/2)gt^2 \Rightarrow (1/2)gt^2 = H/2 \Rightarrow t = \sqrt{H/g}\). Position of A: \(y_A(t) = v_0t - (1/2)gt^2\). At meeting: \(H/2 = v_0t - H/2 \Rightarrow H = v_0t\). Substitute \(t\): \(H = v_0 \sqrt{H/g}\). Squaring both sides gives \(H^2 = v_0^2 (H/g) \Rightarrow v_0^2 = gH\).

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