Question

A projectile is launched at 30° above ground level. what other angle at the same speed will result in the same range?

Answer 1

Answer:

$60^{\circ}$ (assuming that air resistance is negligible.)

Explanation:

Let $g$ denote the gravitational field strength. If air resistance on the projectile is negligible:

• Vertical acceleration of the projectile will be constantly $(-g)$. In other words, $a_{y} = (-g)$.
• Horizontal velocity of the projectile will be constant.

Let $\theta$ denote the angle at which the projectile is launched. Let $v$ denote the initial velocity of the projectile:

• Initial vertical velocity of the projectile will be $u_{y} = v\, \sin(\theta)$.
• Initial horizontal velocity of the projectile will be $u_{x} = v\, \cos(\theta)$.

Also because air resistance is negligible, vertical velocity of the projectile will be $v_{y} = (-u_{y}) = (-v\, \sin(\theta))$ right before the projectile lands. In other words, while the projectile was in the air, the change in vertical velocity would be $(-v\, \sin(\theta)) - (v\, \sin(\theta)) = (-2\, v\, \sin(\theta))$.

Divide the change in velocity by acceleration to find the duration of the flight:

$\begin{aligned}t &= \frac{v_{y} - u_{y}}{a_{y}} \\ &= \frac{(-v\, \sin(\theta)) - (v\, \sin(\theta))}{(-g)} \\ &= \frac{(-2\, v\, \sin(\theta))}{(-g)} \\ &= \frac{2\, v\, \sin(\theta)}{g}\end{aligned}$.

Range measures the horizontal distance that this projectile has travelled. At a constant horizontal velocity of $u_{x} = v\, \cos(\theta)$, this projectile would travel a distance of:

$\begin{aligned}(\text{range}) &= u_{x}\, t \\ &= (v\, \cos(\theta))\left(\frac{2\, v\, \sin(\theta)}{g}\right) \\ &= \frac{2\, v^{2}\, \sin(\theta)\, \cos(\theta)}{g}\end{aligned}$.

Apply the double angle identity $2\, \sin(\theta) \, \cos(\theta) = \sin(2\, \theta)$ to further simplify this expression:

$\begin{aligned}(\text{range}) &= \cdots \\ &= \frac{2\, v^{2}\, \sin(\theta)\, \cos(\theta)}{g} \\ &= \frac{v^{2}\, (2\, \sin(\theta)\, \cos(\theta))}{g} \\ &= \frac{v^{2}\, \sin(2\, \theta)}{g}\end{aligned}$.

Note that in this question, $v^{2}$ and $g$ are both constant. Hence, for another angle of elevation $\hat{\theta}$, the range of the projectile will be the same as long as $\sin(2\, \hat{\theta}) = \sin(2\, \theta)$.

$\sin(2\, \hat{\theta}) = \sin(2\, (30^{\circ}))$.

Since $0^{\circ} \le \hat{\theta} \le 90^{\circ}$, $0^{\circ} \le 2\, \hat{\theta} \le 180^{\circ}$:

$\sin(2\, \hat{\theta}) = \sin(180^{\circ} - 2\, \hat{\theta}) = \sin(2\, (90^{\circ} - \hat{\theta}))$.

Therefore, $\hat{\theta} = 90^{\circ} - \theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$ will ensure that $\sin(2\, \hat{\theta}) = \sin(2\, \theta)$. Launching the projectile at $60^{\circ}$ will reach the same range.

$\begin{aligned}(\text{new range}) &= \frac{v^{2}\, \sin(2\, (60^{\circ}))}{g} \\ &= \frac{v^{2}\, \sin(2\, (30^{\circ}))}{g} = (\text{original range})\end{aligned}$.