Question

Prove that for a given velocity of projection, the horizontal range is the same for two angles of projection q and (90°-q)?​

Answer 1

Answer:

Let the initial velocity of this projectile be $v_0$.

If the object is launched at an angle of projection $q^{\circ}$,

Time of the object in the air:

$\displaystyle \frac{2\;v_0 \sin{q^{\circ}}}{g}$;

Horizontal range:

$\displaystyle \frac{2\;{v_0}^{2}\sin{q^{\circ}}\cdot \cos{q^{\circ}}}{g}$.

If the object is launched at an angle of projection $(90-q)^{\circ}$,

Time of the object in the air:

$\displaystyle \frac{2\;v_0 \sin{(90-q)^{\circ}}}{g}$;

Horizontal range:

$\displaystyle \frac{2\;{v_0}^{2}\sin{(90-q)^{\circ}}\cdot \cos{(90-q)^{\circ}}}{g}$.

However

$\sin{q^{\circ}} = \cos{(90 - q)^{\circ}}$ and $\cos{q} = \sin{(90-q)^{\circ}}$.

Therefore

$\displaystyle \frac{2\;{v_0}^{2}\sin{(90-q)^{\circ}}\cdot \cos{(90-q)^{\circ}}}{g} &= \frac{2\;{v_0}^{2}\cos{q^{\circ}}\cdot \sin{q^{\circ}}}{g} = \frac{2\;{v_0}^{2}\sin{q^{\circ}}\cdot \cos{q^{\circ}}}{g}$.

Hence the horizontal range of the projectile will be the same if the object is launched at an angle of projection $q^{\circ}$ and at $(90-q)^{\circ}$

Explanation:

Let the initial velocity of this projectile be $v_0$. If this projectile is launched at an angle of elevation of $\theta$, the horizontal velocity of this projectile will equal $v_0 \cos{\theta}$. The initial vertical velocity of this projectile will equal $v_0 \sin{\theta}$.

Assume that there's no air resistance on the projectile. The horizontal velocity of the projectile shall be constant. The range of this projectile will be the same as its displacement in the horizontal direction. That is:

Range = Horizontal Displacement = Horizontal Velocity × Time in the Air.

In the vertical direction, gravity causes the object to accelerate downwards at $-g$.

The vertical velocity on the projectile is zero at the vertex of its trajectory. The trajectory of the projectile is symmetric about the vertex. As a result, the projectile reaches the vertex at one-half the total time that the projectile is in the air.

In other words, if $t$ be the total time that the projectile stays in the air, it will reach the vertex of its trajectory at $t/2$. Also, it takes a time of $v_0 \sin{\theta}/g$ for the vertical velocity of the projectile to drop to zero. Since that's also the time it takes for the projectile to reach its vertex,

$\displaystyle \frac{1}{2}t = \frac{v_0 \sin{\theta}}{g}$.

Solve for $t$, the time that the projectile stays in the air:

$\displaystyle t =\frac{2\;v_0 \sin{\theta}}{g}$.

The range of the projectile launched at an angle of elevation $\theta$ and an initial velocity of $v$ will thus equal:

$\displaystyle \frac{2\;v_0 \sin{\theta}}{g} \cdot v_0\cos{\theta} = \frac{2\;{v_0}^{2}\sin{\theta}\cdot \cos{\theta}}{g}$.

The range of a projectile launched at an initial velocity of $v$ will be

• $\displaystyle \frac{2\;{v_0}^{2}\sin{q^{\circ}}\cdot \cos{q^{\circ}}}{g}$ if $q^{\circ}$ is the angle of elevation;
• $\displaystyle \frac{2\;{v_0}^{2}\sin{(90-q)^{\circ}}\cdot \cos{(90-q)^{\circ}}}{g}$ if $(90-q)^{\circ}$ is the angle of elevation.

Note that

$\sin{q^{\circ}} = \cos{(90 - q)^{\circ}}$ and $\cos{q} = \sin{(90-q)^{\circ}}$.

As a result,

$\displaystyle \frac{2\;{v_0}^{2}\sin{(90-q)^{\circ}}\cdot \cos{(90-q)^{\circ}}}{g} &= \frac{2\;{v_0}^{2}\cos{q^{\circ}}\cdot \sin{q^{\circ}}}{g} = \frac{2\;{v_0}^{2}\sin{q^{\circ}}\cdot \cos{q^{\circ}}}{g}$.

Hence the horizontal range of the projectile will be the same for angles of elevation $q^{\circ}$ and $(90 - q)^{\circ}$.