First slide

Projection Under uniform Acceleration

Question

The equation of the path of a projectile is given by $Y = \sqrt{3} x - \frac{x^{2}}{30 \sqrt{3}}$ (x and y are in meters ). The angle of,projection and the range are respectively.

Easy

A

30°, 90m
B

45°, 60m
C

30°, 30m
D

60°, 90m

Solution

$y = \sqrt{3} x - \frac{x^{2}}{30 \sqrt{3}} y = \tan θx - \frac{{gx}^{2}}{2 u^{2} \cos^{2} θ}$
comparing $\tan θ = \sqrt{3} \Rightarrow θ = 60^{\circ}$
$\begin{array}{l} R = \frac{u^{2} \sin 2 θ}{s} = \frac{60^{0} \sqrt{3}}{10} \times \frac{\sqrt{3}}{2} \\ = 90 m \end{array}$

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