Question

A projectile can have same range R for angle x and (90°-x) if t1 and t2 are time of flight then what is the relation between R,t1 and t2 ?

Answer 1

The horizontal range of a projectile is given by,

$R = \frac{ {u}^{2}sin(2x) }{g}$

Since

⇒sin2( 90 - x )= sin( 180-x ) = sinx

Therefore, Range of projectile is same for two angles x and 90 - x.

The Time of flight of a projectile is given by,

$T= \frac{2u \sin( \theta) }{g}$

Given t₁, t₂ are time of flights respectively then,

$t _{1} = \frac{2u \sin( x) }{g} \\ \\ t _{2} = \frac{2u \sin( 90 - x) }{g} = \frac{2u \cos( x) }{g}$

Let us consider,

$R =x t_{1} t_{2}$

Now,

$\frac{ {u}^{2}sin(2x) }{g} = \: x \frac{2u \sin( x) }{g} \times \frac{2u \cos( x) }{g} \\ \\ \sin(2x) = x \times 2 \sin(2x) \times \frac{1}{g} \\ \\ x = \frac{g}{2}$

Therefore, R = 1/2gt₁t₂

Answer 2

Answer:

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