Question

Two balls of equal masses are projected upward simultaneously, one from the ground with speed 50 m/s and other from a 40 m high tower with initial speed 30 m/s. Find the maximum height attained by their centre of mass.

Answer 1

Answer on Question#56938 - Physics – Mechanics | Relativity

Two balls of equal masses are projected upward simultaneously, one from the ground with speed $v_{1} = 50 \, \text{m/s}$ and other from a $h_{0} = 40 \, \text{m}$ high tower with initial speed $v_{2} = 30 \, \text{m/s}$. Find the maximum height attained by their centre of mass.

Solution:

The dependence of the height on time of the first ball is given by

$$h_{1}(t) = h_{0} + v_{1} t - \frac{g t^{2}}{2}$$

The dependence of the height on time of the second ball is given by

$$h_{2}(t) = v_{2} t - \frac{g t^{2}}{2}$$

Since masses of balls are equal, the dependence of the height on time of the centre of mass is given by

$$h_{c}(t) = \frac{h_{1}(t) + h_{2}(t)}{2} = \frac{h_{0}}{2} + \frac{v_{1} + v_{2}}{2} t - \frac{g t^{2}}{2}$$

It reaches maximum height when $h_c'(t) = 0$:

$$\frac{d}{dt} \left( \frac{h_{0}}{2} + \frac{v_{1} + v_{2}}{2} t - \frac{g t^{2}}{2} \right) = 0$$ $$\frac{v_{1} + v_{2}}{2} - g t = 0$$ $$t = \frac{v_{1} + v_{2}}{2 g}$$

Thus the maximum height attained by centre of mass is

$$h_{\max} = h_{c} \left( \frac{v_{1} + v_{2}}{2 g} \right) = \frac{h_{0}}{2} + \frac{1}{g} \left( \frac{v_{1} + v_{2}}{2} \right)^{2} - \frac{g \left( \frac{v_{1} + v_{2}}{2 g} \right)^{2}}{2} = \frac{h_{0}}{2} + \frac{(v_{1} + v_{2})^{2}}{8 g} = \frac{40 \, \text{m}}{2} + \frac{ \left( 50 \, \frac{\text{m}}{\text{s}} + 30 \, \frac{\text{m}}{\text{s}} \right)^{2} }{8 \cdot 9.8 \, \frac{\text{m}}{\text{s}^{2}}} = 101.6 \, \text{m}$$

Answer: 101.6 m.

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