Question

57. The time of flight of a projectile is related to
its horizontal range by the equation gT² = 2R.
The angle of projection is​

Answer 1

Answer:-

$\theta = 45^{\circ}$

Given :-

Time of flight and it's horizontal range is related by equation:-

gT² = 2R

To find :-

The angle of projection.

Solution:-

Let T be the time of flight and R be the range of the projectile.

We know that,

The Time of flight of a projected body is given by :-

$\huge \boxed{T = \dfrac{2uSin\theta}{g}}$

The range of the projected body is given by :-

$\huge \boxed{R = \dfrac{u^2Sin2\theta }{g}}$

Since,

Both are related by equation:-

$gT^2 = 2R$

Put the value of T and R.

→$\mathsf{g \left(\dfrac{2uSin\theta}{g}\right)^2 = 2 \times \dfrac{u^2Sin2\theta }{g}}$

→$\mathsf{ g \times \dfrac{4u^2Sin^2 \theta}{g^2}= \dfrac{2u^2Sin2\theta}{g}}$

• cancelling g on both side.

→$\mathsf{ g \times \dfrac{4u^2Sin^2\theta}{g}= 2u^2Sin2\theta}$

→$\mathsf{ 4u^2 Sin^2 \theta = 2u^2 Sin2\theta }$

→$\mathsf{2u^2 ( 2Sin^2 \theta ) = 2u^2 Sin2 \theta}$

→$\mathsf{\dfrac{2Sin^2 \theta}{Sin2\theta}=\dfrac{2u^2}{2u^2}}$

→$\mathsf{\dfrac{2Sin^2 \theta}{2Sin\theta Cos\theta}= 1 }$

→$\mathsf{ \dfrac{Sin\theta}{Cos\theta}= 1}$

→$\mathsf{Tan\theta = 1 }$

→$\mathsf{Tan\theta = Tan 45^{\circ}}$

→$\mathsf{ \theta = 45^{\circ}}$

hence,

The angle of projection will be $45^{\circ}$