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Chapter 2: Problem 91

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 balls will collide if the initial speed of the ball thrown upwards is \( v_0 = \frac{h_0}{t} - g t\) and they will collide at a height \(h = \frac{1}{2} g t^2\) where \(t = \sqrt{\frac{2 h_0}{g}}\) is the time at which the balls collide.

Step by step solution

01

Equate the heights of the two balls

Set the height of the falling ball equal to the height of the ball thrown upwards. This gives us, \(h_0 - \frac{1}{2} g t^2 = \frac{1}{2} g t^2 + v_0 t\)
02

Simplifying the equation to solve for \(v_0\)

Simplify the equation to derive a condition for \(v_0\) in order for the balls to collide. This will involve rearranging terms and then dividing by \(t\), under the assumption that \(t \neq 0\) since time cannot be zero. This will yield\( v_0 = \frac{h_0}{t} - g t\) where \(v_0\) is the initial speed of the ball thrown upwards.
03

Determine the point of collision

The time at which the collision will take place can be found by solving the quadratic equation: \(g t^2 - 2h_0 = 0\) which yields\(t = \sqrt{\frac{2 h_0}{g}}\) The height at which they collide is then given by \(h = \frac{1}{2} g t^2\) where \(t\) is the time at which the balls collide.

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