Question

At what angle should a projectile have to be fired to cover the maximum horizontal range and why?

Answer 1

We will write the formula for horizontal range and we can easily make out by solving what will be the angle to make the range maximum:

$R = \dfrac{{{v^2}\sin 2\theta }}{g}$, R is the range horizontal range, v is the velocity, g is the gravitational acceleration and $\theta$ is the angle at which the projectile is fired.

Complete answer:

Lets define the term Projectile and Horizontal Range.

A body which is in flight through the atmosphere but is not being propelled by any fuel is called a projectile. The path followed by a projectile is called trajectory. Various examples of projectiles are,

Bombs released from the aeroplane.

A bullet fired from the gun.

A javelin throw by an athlete.

An arrow released from the bow etc .

Horizontal Range: It is the total horizontal distance from the point of projection to the point where the projectile comes back to the plane of projection.

Horizontal range depends on the angle $\theta$ for a given velocity , when sin2$\theta$ will be maximum then range will also be maximum.

$R = \dfrac{{{v^2}\sin 2\theta }}{g}$

$\Rightarrow \sin 2\theta = 1$

(sin2$\theta$ has the maximum value as 1)$\theta$

$\Rightarrow \sin 2\theta = \sin {90^0}$ (because sin90o is 1)

$\Rightarrow 2\theta = {90^0}$

$\Rightarrow \theta = \dfrac{{{{90}^0}}}{2}$

$\Rightarrow \theta = {45^0}$

We got the angle as ${45^0}$ at which the horizontal range will be maximum because when $\sin (2 \times {45^0}) = \sin {90^0}$, we know $sin{90^0}$=1.

Note: We have considered the formula of horizontal range keeping in mind that there is no resistance due to air, the effect due to curvature of earth is negligible, the effect due to rotation of earth is negligible, for all points of trajectory the value of gravitational acceleration is constant in magnitude and direction.