Question

A particle is projected from the ground with an initial speed of v at an angle θ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

Moderate

Solution

The average velocity is given by
$v_{av} = \frac{displacement}{time}$
The situation of the particle at the highest point is shown in fig. (4).
We know that
Maximum height, $H = \frac{v^{2} \sin^{2} θ}{2 g}$ ….(i)
Range, $R = \frac{v^{2} \sin 2 θ}{g}$ ….(ii)
and Time of flight, $T = \frac{2 vsin θ}{g}$ ….. (iii)
Now using triangle OPQ, we have
$v_{av} = \frac{OP}{(T / 2)} = \frac{2 (OP)}{T}$
or $v_{av} = \frac{2 {(H^{2} + R^{2} / 4)}^{1 / 2}}{T}$ …(iv)
substituting the values in eq. (iv) from eqs. (i), [ii) and (iii), we get
$v_{av} = \frac{v}{2} \sqrt{(1 + 3 \cos^{2} θ)}$

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