Question

a projectile is projected with some velocity at an angle theta from the horizontal so that its range is R and time of flight is T then
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Answer 1

Answer:

$R=\frac{gT^2}{2\:tan\theta}$

option (3) is correct

Explanation:

Formula used:

Time of flight, $T=\frac{2u\:sin\theta}{g}$

Horizontal range, $R=\frac{u^2\:sin2\theta}{g}$

Now,

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

squaring on both sides

$T^2=\frac{4u^2\:sin^2\theta}{g^2}$........................(1)

$R=\frac{u^2\:sin2\theta}{g}$

$R=\frac{u^2\:2\:sin\theta\:cos\theta}{g}$

$R=\frac{g}{2\:sin\theta}(\frac{u^2\:4\:sin^2\theta\:cos\theta}{g^2})$

$R=\frac{g\:cos\theta}{2\:sin\theta}(\frac{4u^2\:sin^2\theta}{g^2})$

$R=\frac{g\:cos\theta}{2\:sin\theta}(T^2)$ (using (1))

$R=\frac{gT^2\:cos\theta}{2\:sin\theta}$

$R=\frac{gT^2}{2\frac{sin\theta}{cos\theta}}$

$R=\frac{gT^2}{2\:tan\theta}$