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If a projectile is launched from a height greater than zero and landed to a height equal to zero, is the optimum launch angle that gives the greatest horizontal range still $45$ degrees or not?

I know that if the projectile is landed to a height not equal to the launch height, the formula $$ R = \frac{v_0^2 \sin2\theta}{g} $$ that maximizes the range when the angle is $45$ degrees is not already applicable. But is this an argument to say that $45$ degrees is not the optimum launch angle for an object launched above the ground and landed to the ground? If $45$ degrees is not the optimum launch angle, in this case, is the greater the angle the lesser the projectile's horizontal range?

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5 Answers 5

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If you say that the projectile is located at a height, then you cannot say 45 degrees is the optimum launch angle (this would however be correct on a flat plane with the projectile elevation $h = 0$ units). If however, the projectile is launched at a height $h$ above the plane, the optimum angle will be equal to a function $f(v_0,h)$.

You can easily calculate the time take for the projectile to travel from $A$ to $B$. For the time taken from $B$ to $C$, think about the energy possessed by the projectile at the two points. Also remember that the $x$ component of the velocity of the projectile does not change throughout the flight. Calculate the total range in the $x$ direction and remember that largest distance travelled is when $\frac{dR}{d\theta}=0$.

Simplify the equation formed and you'll get the optimum angle as a function $f(v_0,h)$.Parabola not to scale!

For time taken from A to C, $$v_y = -v_0sin(\theta)$$ $$a=g$$ Using equations of motion and considering downward direction to be positive, $$h=v_{y}t+\frac{1}{2}at^{2}$$ Solving for $t$ gives $$t=\frac{v_{0}\sin\left(\theta\right)}{g}\left(1+\sqrt{1+\frac{2gh}{v_{0}^{2}\sin^{2}\left(\theta\right)}}\right)$$

X component of velocity ($v_0\cos\theta$) does not change throughout flight.

$$R = v_0\cos(\theta)t = \frac{v_0^2\sin2\theta}{2g}\left(1+\sqrt{1+\frac{2gh}{v_0^2\sin^2\theta}}\right)$$

Now $\frac{dR}{d\theta}=0$ for max range. Implicitly differentiating is much helpful as outlined here.

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  • $\begingroup$ If you say that the projectile is located at a height, then you cannot say 45 degrees is the optimum launch angle I'm pretty sure you are wrong about that. Even at $y=h$ the maximum horizontal distance travelled remains for $45^{\circ}$. IMO. $\endgroup$
    – Gert
    Dec 18, 2020 at 12:41
  • $\begingroup$ I think you meant: $\theta = \frac{1}{2}\cos^{-1}\Big(\frac{gh}{v_0^2+gh}\Big)$. But you don't really show how you arrived at this simple relationship? $\endgroup$
    – Gert
    Dec 18, 2020 at 17:38
  • $\begingroup$ For me you can 'break' the rules. ;-) $\endgroup$
    – Gert
    Dec 19, 2020 at 13:06
  • $\begingroup$ But actually I will try R.W.Bird's method. $\endgroup$
    – Gert
    Dec 19, 2020 at 13:20
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    $\begingroup$ Mastermind 817: Your expression for $t_{BC}$ is missing a g, the expression under the square root in R is not dimensionless, and finding that derivative must have been fun. $\endgroup$
    – R.W. Bird
    Dec 19, 2020 at 16:48
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I'll try to give a qualitative answer without the need for math.

While the 45° angle gives the maximum distance for same height, this has to be adjusted for height differences, resulting in a flatter optimum angle. Why?

We know that the projectile follows a parabola, meaning that on its downward path it'll pass through launch height under the same angle as it was launched.

The path can be divided into two parts, the above-launch and the below-launch part.

Let's imagine an angle slightly flatter, e.g. 44°. The above-launch part will still make nearly the same distance as the 45° launch (close to an optimum, things change slowly), but the continuation (the below-launch part) will make more distance because of the flatter angle.

An angle steeper than 45° cannot result in a greater overall distance, as then both parts make a shorter distance (the above-launch part because we are away from its 45° optimum, and the below-lauch part because of the steeper downward angle).

The optimum will surely be at some positive, upward angle, as launching with a downward component will surely be worse (it reduces both horizontal speed as well as flight time when compared to a horizontal launch).

So the question remains where between 0° and 45° is the optimum. This can only be answered exactly using math, at some angle where the loss in above-launch distance is no longer compensated by the gain in below-launch distance.

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The angle for maximum range from a height, h, is not 45 degrees. To find it start with the component equations: x = $v_o$ cos(θ) t and y = h + $v_o$ sin(θ) t – (1/2)g$t^2$ = 0. Solve the x equation for cos(θ) and the y equation for sin(θ). Then $sin(θ)^2$ + $cos(θ)^2$ = 1. This leads to a quadratic equation in $t^2$ which gives two positive values for t. These correspond to the two possible angles for hitting a target a known distance down range. As you approach the maximum range, the angles (and times) converge. At this point the square root in the quadratic equals zero. You can set it equal to zero and solve for $x^2$. With the square root at zero the quadratic gives $t^2$. Combine these in the x equation to get cos(θ).

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If $h>0$ then, as other answers have pointed out, the time of flight $t$ is the positive root of a quadratic, and the range $R=v_xt$ is

$$R(\theta) = \frac{v_0^2 sin(2\theta)}{2g}\left(1+\sqrt{1+\frac{4gh}{v_0^2 \cos^2 \theta}} \right)$$

where $\theta$ is the angle between the launch angle and the vertical. Finding the value of $\theta$ which maximises $R$ is difficult, but if $\frac{gh}{v_0^2} << 1$ we can approximate $R(\theta)$ by

$$R(\theta) \approx \frac{v_0^2\sin(2\theta)}{g}+2 \tan (\theta) h$$

and then

$$\frac{dR}{d\theta} \approx \frac{2v_0^2\cos(2\theta)}{g}+2 \sec^2 (\theta) h$$

When $0 < \theta < 45^o$ we have $\frac{dR}{d \theta}>0$ and at $\theta = 45^o$ we have $\frac{dR}{d \theta} \approx 4h > 0$, so to maximise $R(\theta)$ we need to make $\theta$ greater than $45^o$ i.e. the optimal trajectory is shallower than it is when $h=0$.

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  • $\begingroup$ How did you get the first equation? $\endgroup$
    – user282164
    Dec 20, 2020 at 12:11
  • $\begingroup$ But is the equation for range as a function of launch angle that I commented in answer above correct? $\endgroup$
    – user282164
    Dec 20, 2020 at 12:14
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    $\begingroup$ @JordanG The first equation comes from finding the positive root of the quadratic equation satisfied by the time of flight $t$ and then using this value of $t$ in $R=v_xt$. The equation you gave for $R$ is only correct when $h=0$. It is not correct when $h>0$. $\endgroup$
    – gandalf61
    Dec 20, 2020 at 12:36
  • $\begingroup$ There's one comment above by Mastermind817 for the range equation which looks similar to yours but are really not the same. I've followed his steps and it seems that I've got it. But I'm really unsure about yours, if you did something more correct in deriving that equation. $\endgroup$
    – user282164
    Dec 20, 2020 at 12:54
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I know that if the projectile is landed to a height not equal to the launch height, the formula that maximizes the range when the angle is 45 degrees is not already applicable. But is this an argument to say that 45 degrees is not the optimum launch angle for an object launched above the ground and landed to the ground?

Trajectory

The angle providing the greatest $R$ remains $45^{\circ}$. But launched at that angle from an elevated ($y>0$) launch position actually increases $R$ somewhat, with respect to launching from $y=0$.

So if you wanted to hit the same target from an elevated launch position, you'd have to adjust the angle or the initial velocity.


Semi-mathematically we can show the horizontal distance travelled is given as follows.

The time spent airborne for the case $y=0$ is:

$$t_0=\frac{2v_0\sin\alpha}{g}$$

And because vertical and horizontal velocities are independent of each other (Galilean invariance), the distance travelled horizontally is:

$$R_0=t_0v_0\cos\alpha$$

But in the case that $y>0$ then the time $t_1$ spent airborne is longer and so:

$$R_1=t_1 v_0\cos\alpha>R_0$$

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  • $\begingroup$ $$ R = \frac{-b \sqrt{b^2 - 4ac}}{2a} $$ where $a=\frac{g}{2v_0^2 cos^2 \theta} $ , $b=-tan \theta $ , $c=-y_0$ $\endgroup$
    – user282164
    Dec 18, 2020 at 13:04
  • $\begingroup$ is that the formula for range if the initial height is above the ground? $\endgroup$
    – user282164
    Dec 18, 2020 at 13:05
  • $\begingroup$ I used the formula $R= v_0xt$ and $y=y_0 + v_0yt - \frac{1}{2} gt^2$ $\endgroup$
    – user282164
    Dec 18, 2020 at 13:06
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    $\begingroup$ If $h>0$ the launch angle that maximises $R$ is not $45^o$ - see my answer below. $\endgroup$
    – gandalf61
    Dec 18, 2020 at 16:29