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Chapter 12: Problem 32

Ball \(A\) is thrown vertically upwards with a velocity of \(v_{0} .\) Ball \(B\) is thrown upwards from the same point with the same velocity \(t\) seconds later. Determine the elapsed time \(t<2 v_{0} / g\) from the instant ball \(A\) is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

Short Answer

Expert verified
The elapsed time from the moment ball \(A\) is thrown until the balls pass each other is \(2t\) seconds. The velocities of the balls at this point can be calculated using the equation \(v = v_0 - g*t\), where \(t\) for ball \(A\) is \(2t\) and for ball \(B\) is \(t\).

Step by step solution

01

Determine the Equations of Motion for Both Balls

The equation of motion for an object thrown vertically upwards is given by \(h = v_0*t - 0.5*g*t^2\), where \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity, \(h\) is the height at time \(t\), and \(t\) is the time passed since the object was thrown. Therefore, for ball \(A\), its height at time \(t\) is \(h_A = v_0*t - 0.5*g*t^2\). For ball \(B\), which is thrown \(t\) seconds later, its height at time \(t'\) is \(h_B = v_0*t' - 0.5*g*t'^2\), where \(t' = t - t\).
02

Determine When the Balls Will Pass Each Other

The balls will pass each other when \(h_A = h_B\), i.e., when they reach the same height. Solving the equation \(h_A = h_B\) for \(t'\), we have \(t = t + t\). The elapsed time from when \(A\) is thrown should therefore be \(t = 2t\).
03

Calculate the Velocities

The velocity of an object thrown vertically upwards is given by \(v = v_0 - g*t\). Therefore, the velocity of ball \(A\) at time \(t\) is \(v_A = v_0 - g*t\) and that of ball \(B\) at time \(t'\) is \(v_B = v_0 - g*t'\), where \(t' = t - t\). Since we have found that \(t = 2t\), we can substitute this value into the velocity equations to find the velocities of the balls.

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