Physics

Physics

Problem

A projectile must pass through a point that is 30 m30\,\text{m} horizontally and 10 m10\,\text{m} vertically above its launch point exactly 2.0 s2.0\,\text{s} after launch. Neglect air resistance and take g=9.8 m ⁣⋅ ⁣s−2g=9.8\,\text{m}\!\cdot\!\text{s}^{-2}. What launch angle relative to the horizontal will satisfy this condition? a) 59∘59^{\circ} b) 45∘45^{\circ} c) 18∘18^{\circ} d) 33∘33^{\circ}
Intuitive explanation

The key is to use the kinematic equations for horizontal and vertical motion independently. Since you know the displacement in both directions and the time, you can extract the initial velocity components—then the launch angle follows from their ratio. Horizontal motion is uniform, while vertical motion involves constant acceleration, so set up two equations and solve for the angle that makes both work simultaneously.

Intuitive explanation

The key is to use the kinematic equations for horizontal and vertical motion independently. Since you know the displacement in both directions and the time, you can extract the initial velocity components—then the launch angle follows from their ratio. Horizontal motion is uniform, while vertical motion involves constant acceleration, so set up two equations and solve for the angle that makes both work simultaneously.

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