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A ball is projected vertically upwards from ground level with speed u1. At the point when this ball is at its maximum height, another ball is launched with initial velocity u2. The two balls fall back on the ground at the same time and do not collide at any moment. Find the ratio of u1:u2.

I can only solve this problem by choosing certain values of u1, such as 10, which gives me a ratio of 2:1. If you could find a general way to find the answer without having to choose initial values for u1, I would be very thankful.

AOD
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1 Answers1

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The speed and position of ball 1 is given by the following equations: \begin{eqnarray} v_{1}(t) = -gt + u_{1} \quad \mbox{and} \quad y_{1}(t) = -\frac{gt^{2}}{2}+u_{1}t \end{eqnarray} Thus, the first ball attains its maximum height at a time $t_{M}$ given by: $$t_{M} = \frac{u_{1}}{g}$$

Now, in order to study the motion of both balls at the same time, you have to consider that the second ball was launched after the first one attained its maximum height, so the equations of motion of the second ball should be translated in time by an amount $t_{M}$. Thus, the equation of motion of the second ball is given by: \begin{eqnarray} y_{2}(t) = -\frac{g(t-t_{M})^{2}}{2}+u_{2}(t-t_{M}) \end{eqnarray} The first ball comes back to the ground at $$T_{1}= \frac{2u_{1}}{g}$$ and the second ball comes back to the ground at: $$T_{2}= t_{M}+\frac{2u_{2}}{g}$$ But, by hypothesis, $T_{1}=T_{2}$, so that: $$\frac{2u_{1}}{g} = \frac{u_{1}}{g}+\frac{2u_{2}}{g} \Rightarrow \frac{u_{1}}{u_{2}} = 2.$$

IamWill
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