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Chapter 2: Problem 98

A ball is dropped from rest at a height \(h_{0}\) above the ground. At the same instant, a second ball is launched with speed \(v_{0}\) straight up from the ground, at a point directly below where the other ball is dropped. (a) Find a condition on \(v_{0}\) such that the two balls will collide in mid-air. (b) Find an expression for the height at which they collide.

Short Answer

Expert verified
The condition for collision is that the initial velocity \(v_0\) must be equal to the height divided by the time (\(v_0 = h_0/t\)). The height at which they collide is \(h_2 = h_0 - 1/2gt^2\).

Step by step solution

01

Establish equations of motion for both balls

Let's start by considering the situation when the two balls meet. The time elapsed since the start is the same for both balls. Thus, we can write the position of the first ball (dropped from a height \(h_0\)) and the second ball separately. The equation for the first ball dropped from rest is: \(h_1 = h_0 - 1/2gt^2\). Here, g is the acceleration due to gravity and t is the time. The equation for the second ball (launched upward) is: \(h_2 = v_0t - 1/2gt^2\).
02

Determine the condition for collision

The two balls will collide when they are at the same height at the same time. Therefore, we equate the two equations: \(h_0 - 1/2gt^2 = v_0t - 1/2gt^2\). Solving for \(v_0\) we find that the condition for \(v_0\) is: \(v_0 = h_0/t\).
03

Calculate the height of collision

Substitute the value for \(v_0\) from Step 2 into the equation for \(h_2\): \(h_2 = h_0 - 1/2gt^2 = h_0t - 1/2gt^2\). The height at which they collide is thus: \(h_2 = h_0 - 1/2gt^2\).

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