Ball Collision Height – Rankers Physics

Relative Motion in One Dimension: Practice Problem & Solution

Two balls are thrown simultaneously, (A) vertically upwards with a speed of \(20\text{ m/s}\) from the ground and (B) vertically downwards from a height of \(80\text{ m}\) with the same speed and along the same line of motion. At which point will the balls collide? (take \(g = 10\text{ m/s}^2\)
15 m above the ground
15 m below from the top of the tower
20 m above from the ground
20 m below from the top of the tower

Solution Explained:

To solve this problem, we apply the core principles of Relative Motion in One Dimension. Understanding the underlying formula is key to arriving at the correct answer below:

Let \(y_A = 20t - 5t^2\) and \(y_B = 80 - 20t - 5t^2\). Equating them for collision: \(20t - 5t^2 = 80 - 20t - 5t^2\). This simplifies to \(40t = 80\), so \(t = 2\) s. The collision height is \(y_A(2) = 20(2) - 5(2)^2 = 40 - 20 = 20\) m above the ground.

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