First slide

Projection Under uniform Acceleration

Question

Two particles are projected from the ground simultaneously with speeds 20 m/s and $20 / \sqrt{3} m / s$ at an angles 30° and 60° with the horizontal in the same direction. The maximum distance between them till both of them strike the ground is approximately (g =10 m /s²)

Moderate

A

23.1 m
B

16.4 m
C

30.2 m
D

10.4 m

Solution

Components of velocities of both the particles in vertical direction are equal. Therefore, their times of flight are equal and their relative motion is in horizontal direction only. Thus, the maximum distance between them is equal to the difference in their horizontal ranges.
$\begin{array}{l} R_{1} = \frac{(20)^{2} \sin 60^{\circ}}{g} = \frac{400}{10} \frac{\sqrt{3}}{2} \\ = 34 .64 m \\ R_{2} = \frac{(20 / \sqrt{3})^{2} \sin 120^{\circ}}{g} m \\ = \frac{200}{\sqrt{3} g} = 11 .55 m \\ ∴ S_{max .} = R_{1} - R_{2} = 23 .09 m \end{array}$

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