# Projectile motion - calculate landing angle

A projectile is launched at a $45^\circ$ angle, aiming for a target at a distance of 15 feet away from, and $2$ feet below the starting position.

I'm looking for equations to determine:

• The initial velocity of the projectile required to hit the target
• The angle at which the projectile hits the target

I'm particularly interested in learning if there are equations to solve this, as I'm interested in plugging in different distance/height offsets in the future.

We can first find the angle since the initial and final velocities are not given, while the initial displacement are given as X = Y = 0 and the final displacement as X = 4.57 m and Y = 0.61 m.

The goal is to eliminate the initial unknown velocities where $$v_x = v_i\cos(\theta)$$ $$v_y = v_i\sin(\theta)$$ To do this we use equations $$x = v_i\cos(\theta)t$$ $$y = v_i\sin(\theta)t-\frac12gt^2$$ Substituting for $v_i$ we get have $$y = \tan(\theta)x-\frac 12 g\left(\frac x{v_i\cos\theta}\right)^2$$ We can use this equation to solve for $\theta$ which is the angle at which the target is hit.

Given $\theta$ we can find $v_i$ using $$y = v_i\sin(\theta)t-\frac12gt^2$$

• Thanks very much, but I don't see how to rearrange the substituted equation (y=tan...) to give the angle (Apologies, my math is pretty rusty!). – X__ Dec 14 '15 at 5:45
• The equation y = is a transcendental one which must be solved iteratively for theta, given values for x and y – user16622 May 27 '16 at 8:58

\begin{aligned} x &= \frac{v}{\sqrt{2}} t \\ y & = \frac{v}{\sqrt{2}} t - \frac{1}{2} g t^2 \end{aligned}

To hit the target at $x$ you need $t =\frac{x \sqrt{2}}{v}$ time. At this time the vertical position is

$$y = \frac{v}{\sqrt{2}} \left( \frac{x\sqrt{2}}{v} \right) - \frac{1}{2} g \left( \frac{x \sqrt{2}}{v} \right)^2 = x - \frac{g x^2}{v^2}$$

Solve the above for $v$ the initial velocity.

Once you have the initial velocity calculate the slope by calculating the velocity components in the x and y directions as a function of time.

\begin{align} \dot{x}(t) &= \frac{v}{\sqrt{2}} \\ \dot{y} (t)&= \frac{v}{\sqrt{2}} - g t \end{align}

and $$\mbox{(slope)} = \frac{\dot{y}}{\dot{x}} = 1 - \frac{g t \sqrt{2}}{v}$$

• Just for a check I get the angle to be below 60° and above 50° in a negative direction. – ja72 Dec 14 '15 at 2:53
• I'm getting -55, so that seems to fit your range - thanks! Just to check, is the square root of 2 part for the formulae above the offset for the height? That is, if the target was 4 feet below the starting location, would I change these formulas to take the square root of 4 where it was previously the square root of 2? – X__ Dec 14 '15 at 4:45
• the $\sqrt{2}$ comes from the 45° angle of launch. Did you plug in the final time $t=\frac{x\sqrt{2}}{v}$ and then the launch speed $v$ to get the angle. My result was between -50° and -52° actually. Also remember to $\theta=\arctan\mbox{(slope)}$ – ja72 Dec 14 '15 at 13:00
• Still not getting there - maybe we can check where I'm making my error. I get v as 20.64404, t = 1.03, x(t) = 14.60533 y(t) = -1693.44 – X__ Dec 14 '15 at 17:29
• I might have screwed up the units. I blame on the idiotic feet-inches debacle. For the target $(x,y)=(15,-2)$ I get \begin{aligned} t & =\sqrt{ \frac{2(x-y)}{g}} & v & = x \sqrt{\frac{g}{x-y}} \end{aligned} – ja72 Dec 14 '15 at 18:00