Question

A body is projected with same speed at two different angles cover the same horizontal distance r.If t1 and t2 are the time of flights then r is equal to?

Answer 1

Answer:

hopefully u have got the ans

Answer 2

Answer:

$r = \frac{( g \times t1 \times t2 )}{2}$

Explanation:

For two bodies the two bodies to have the same displacement but different angles, the phase difference or difference between the angles should be 90°.

Let the time of flight for first body be

$\frac{2u \sin(θ) }{g}$

Let the time of flight for second body be

$\frac{2u \sin(90 - θ) }{g}$

Then,

$t1 \times t2 = \frac{4 {u}^{2} \sin(θ) \cos(θ) }{g}$

We know that the range R is given by

$r = \frac{{u}^{2} \sin(2θ)}{g}$

Rearranging a little bit we get

$r = \frac{( g \times t1 \times t2 )}{2}$

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