Question

Derive and solve the equation of motion of a particle, in a uniform gravita-
tional field, projected with initial horizontal velocity v 0 at a height h.

Answer 1

Answer:

Hey !!

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Let us consider a ball projected at an angle θ with respect to horizontal x-axis with the initial velocity u .

The point O is called the point of projection, θ is the angle of projection and OB = horizontal range. The total time taken by the particle from reaching O to B is called the time of flight.

Now,

(a). The total time of flight is

Resultant displacement is zero in Vertical direction.

Therefore, by using equation of motion

$s = ut - \frac{1}{2} {gt}^{2}$

$gt = 2 \sinθ$

$t = \frac{2sinθ}{g}$

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$R=ucosθ× \frac{2usinθ}{g}$

$R= \frac{ {u}^{2}sin2θ }{g}$

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${v}^{2} - {u}^{2} = 2as$

$0 = {u}^{2} {sin}^{2} </p><p>θ−2gH$

$H= \frac{ {u}^{2} {sin}^{2} θ }{2g}$

Hence, this is the required solution.

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Explanation:

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