Question

A particle is projected vertically upwards. After a time t, another particle is sent up from the same point with the same velocity and meets the first at a height h during the downward flight of the first. Find the velocity of projection.

Answer 1

Answer on Question #67215-Physics-Solid State Physics

A particle is projected vertically upwards. After a time $t$, another particle is sent up from the same point with the same velocity and meets the first at a height $h$ during the downward flight of the first. Find the velocity of projection.

Solution

$$y = v T - \frac {g T ^ {2}}{2}$$ $$y = v (T - t) - \frac {g (T - t) ^ {2}}{2}.$$

A height h:

$$v T - \frac {g T ^ {2}}{2} = h \rightarrow T = \frac {v \pm \sqrt {v ^ {2} - 4 \left(\frac {g}{2}\right) h}}{2 \left(\frac {g}{2}\right)}$$

It meets the first at a height $h$ during the downward flight of the first:

$$T = \frac {v + \sqrt {v ^ {2} - 2 g h}}{g}.$$

So,

$$v \left(\frac {v + \sqrt {v ^ {2} - 2 g h}}{g} - t\right) - \frac {g \left(\frac {v + \sqrt {v ^ {2} - 2 g h}}{g} - t\right) ^ {2}}{2} = h$$ $$2 \sqrt {v ^ {2} - 2 g h} = g t$$ $$v ^ {2} = 2 g h + \left(\frac {g t}{2}\right) ^ {2}$$

The velocity of projection is

$$v = \sqrt {2 g h + \left(\frac {g t}{2}\right) ^ {2}}$$

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