Projectile targets

Objective. Solve impact and interception constraints.

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2. Worked example
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Learn: the key idea

Projectile target problems set up equations from the horizontal and vertical displacement formulas to find conditions, such as launch angle or speed, that make a projectile hit a specified point (x, y), modelling interception or impact constraints. Typically the trajectory equation y = x tan θ − gx²/(2u²cos²θ) is used directly, substituting the target coordinates and solving for the unknown, often producing a quadratic in tan θ after using the identity sec²θ = 1 + tan²θ. For example, finding the angle to hit a target 50 m away and 5 m high requires solving such a quadratic for tan θ, then choosing the physically sensible solution.

Worked example

A ball is thrown from ground level and must pass through a point 20 m away at height 0 (landing point), with speed u = 20 m/s. Find the launch angle θ using the range equation, taking g = 10 m/s².

1 $Range equation: R = u²sin2θ \div g$
2 $20 = 400 sin2θ \div 10, so 20 = 40 sin2θ$
3 $sin2θ = 0.5, so 2θ = 30° or 150°$
4 $θ = 15° or θ = 75°$

Answer: θ = 15° or θ = 75°

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Common mistake. Students forget there can be two valid launch angles (a low, flat trajectory and a high, lobbed one) satisfying the same target, and only report one root of the quadratic. Always check the quadratic in tan θ for both possible solutions.

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