Perpendicular projectiles

Classical Mechanics Level 2

Two particles move in a uniform gravitational field with an acceleration g g . At the initial moment the particles were located at one point and moved with velocities v 1 = 3 m/s v_1 =3 \text{ m/s} and v 2 = 4 m/s v_2 =4 \text{ m/s} horizontally in opposite directions.

Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular


The answer is 2.5.

Vishal S by Vishal S , 21, India

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1 solution

Steven Chase
Jul 12, 2017

( v x 1 , v y 1 ) = ( 3 , − g t ) ( v x 2 , v y 2 ) = ( − 4 , − g t ) (v_{x1},v_{y1}) = (3, -gt) \\ (v_{x2},v_{y2}) = (-4, -gt)

When the velocities are perpendicular, the v x v y \large{\frac{v_x}{v_y}} ratios are the negative reciprocals of each other.

3 − g t f = − g t f 4 ⟹ g 2 t f 2 = 12 ⟹ t f = 12 g 2 \frac{3}{-gt_f} = -\frac{gt_f}{4} \implies g^2 t_f^2 = 12 \implies t_f = \sqrt{\frac{12}{g^2}}

The y y (vertical) velocities are always the same, so they do not contribute to the relative displacement. The relative horizontal speed is 7. Therefore, the distance between the particles is 7 t f 7 t_f when the vector velocities are perpendicular.

Δ = 7 t f = 7 12 g 2 \Delta = 7 t_f = 7 \sqrt{\frac{12}{g^2}}

Plugging in 9.8 9.8 for g g gives Δ ≈ 2.47 \Delta \approx 2.47

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