Question

Air drop—inverse problem A plane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 m must drop an emergency packet on a target on the ground. The trajectory of the packet is given by
x = 100t, y = −4.9t² + 4000, t ≥ 0
where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the target?

Accepted Answer

Identify the variables and given information: The horizontal position is given by $$x = 100t$$, and the vertical position is $$y = -4.9t$$^{2}$$ + 4000. $$The target is on the ground, so the vertical position at impact is $$y = 0. $$Set the vertical position equation equal to zero to find the time $$t$$ when the packet hits the ground: $$-4.9t^{2} + 4000 = 0$. Solve the quadratic equation for t$ to find the time of impact. Since t$$ \geq 0$$, take the positive root of the equation. Use the time of impact t$ in the horizontal position equation x = 100t$ to find the horizontal distance traveled by the packet from the release point to the target. The horizontal distance found is how far before the target the packet must be released, so the plane should release the packet this many meters before reaching the target.

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Key Concepts

Parametric Equations of Motion

Solving for Time of Flight

Inverse Problem in Projectile Motion