Hitting a Target: The Two Possible Angles

A target sits at $(x_t, y_t)$. You have a fixed launch speed $v_0$. What angle should you fire? Surprisingly, there are usually two valid answers — a flat shot and a high lob — both of which pass through the target. Solving this requires turning the trajectory equation into a quadratic in $\tan\theta$.

The Physics

Start from the trajectory equation:

$y = x\tan\theta - \dfrac{g\,x^{2}}{2 v_0^{2}\cos^{2}\theta}$

The trick is to use the identity $\sec^{2}\theta = 1 + \tan^{2}\theta$, which lets us rewrite $1/\cos^{2}\theta = 1 + \tan^{2}\theta$. After substitution and rearrangement, the equation becomes a quadratic in $\tan\theta$:

$\dfrac{g\,x_t^{2}}{2 v_0^{2}}\,\tan^{2}\theta \;-\; x_t\tan\theta \;+\; \left(y_t + \dfrac{g\,x_t^{2}}{2 v_0^{2}}\right) = 0$

The two solutions (when the discriminant is non-negative) give a low-angle and a high-angle trajectory.

Step-by-Step Solution

Given:

• Launch speed: $v_0 = 25\;\text{m/s}$
• Target position: $(x_t, y_t) = (50, 10)\;\text{m}$
• Gravity: $g = 9.81\;\text{m/s}^{2}$

Find: The two launch angles (if any) that hit the target.

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Step 1 — Compute the convenient constant $A = \dfrac{g\,x_t^{2}}{2 v_0^{2}}$.

$A = \dfrac{9.81 \times (50)^{2}}{2 \times (25)^{2}} = \dfrac{9.81 \times 2500}{1250} = \dfrac{24525}{1250} \approx 19.620$

Step 2 — Set up the quadratic in $u = \tan\theta$.

The quadratic is $A\,u^{2} - x_t\,u + (y_t + A) = 0$:

$19.620\,u^{2} - 50\,u + (10 + 19.620) = 0$

$19.620\,u^{2} - 50\,u + 29.620 = 0$

So we have the standard quadratic $a\,u^{2} + b\,u + c = 0$ with:

$a = 19.620 \qquad b = -50 \qquad c = 29.620$

Step 3 — Compute the discriminant $\Delta = b^{2} - 4ac$.

$\Delta = (-50)^{2} - 4(19.620)(29.620)$

$= 2500 - 4 \times 581.14$

$= 2500 - 2324.58$

$\Delta \approx 175.42$

Since $\Delta > 0$, the target is reachable and there are two real solutions.

Step 4 — Apply the quadratic formula.

$u = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{50 \pm \sqrt{175.42}}{2 \times 19.620} = \dfrac{50 \pm 13.244}{39.240}$

The two roots are:

$u_{\text{low}} = \dfrac{50 - 13.244}{39.240} = \dfrac{36.756}{39.240} \approx 0.9367$

$u_{\text{high}} = \dfrac{50 + 13.244}{39.240} = \dfrac{63.244}{39.240} \approx 1.6117$

Step 5 — Take the arctangent of each root.

$\theta_{\text{low}} = \arctan(0.9367) \approx 43.13°$

$\theta_{\text{high}} = \arctan(1.6117) \approx 58.18°$

Step 6 — Verify by plugging $\theta_{\text{low}}$ back in.

Quick check at $\theta = 43.13°$:

$v_{0x} = 25\cos(43.13°) \approx 18.25\;\text{m/s}$

$v_{0y} = 25\sin(43.13°) \approx 17.08\;\text{m/s}$

$t = 50/18.25 \approx 2.740\;\text{s}$

$y = (17.08)(2.740) - \tfrac{1}{2}(9.81)(2.740)^{2} = 46.80 - 36.81 \approx 9.99\;\text{m} \;\;\checkmark$

Just about exactly 10 m. ✓

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Answer: Both launch angles $\theta_{\text{low}} \approx 43.13°$ and $\theta_{\text{high}} \approx 58.18°$ (measured from the horizontal) will land a projectile at the target $(50, 10)$ when the launch speed is 25 m/s. The low-angle shot arrives faster and flatter; the high-angle shot lobs over and arrives slower from a steeper angle. Both are mathematically valid in vacuum.

When Is the Target Unreachable?

The discriminant must be non-negative. The minimum launch speed for a target at $(x_t, y_t)$ is:

$v_{\min}^{2} = g\bigl(y_t + \sqrt{x_t^{2} + y_t^{2}}\bigr)$

For $(50, 10)$: $v_{\min}^{2} = 9.81(10 + \sqrt{2600}) = 9.81(10 + 50.99) = 9.81 \times 60.99 \approx 598.3$, so $v_{\min} \approx 24.46\;\text{m/s}$. Our $v_0 = 25\;\text{m/s}$ is just barely enough — that's why the two angles are close to each other (43° and 58°) instead of widely separated.

Try It

• Drag the target X / Y sliders — see both solution arcs update live.
• If the target glows red, no angle reaches it. Increase v₀ until both arcs appear.
• Toggle the show low / show high switches to compare them individually.
• At the minimum reachable speed, both solutions converge to a single angle of $\theta = 45° + \arctan(y_t/x_t)/2$.