First slide

Projection Under uniform Acceleration

Question

The equation of motion of a projectile is $y = 12 x - \frac{3}{4} x^{2}$ The horizontal component of velocity is 3 ms^-1. What is the range of the projectile?

Easy

A

18 m
B

16 m
C

12 m
D

21.6 m

Solution

$Given y = 12 x - \frac{3}{4} x^{2}, u_{x} = 3 {ms}^{- 1} v_{y} = \frac{dy}{dt} = 12 \frac{dx}{dt} - \frac{3}{2} x \frac{dx}{dt} At x = 0, v_{y} = u_{y} = 12 \frac{dx}{dt} - 12 u_{x} = 12 \times 3 = 36 {ms}^{- 1} a_{y} = \frac{d}{dt} (\frac{dy}{dt}) = 12 \frac{d^{2} x}{{dt}^{2}} - \frac{3}{2} [{(\frac{dx}{dt})}^{2} + x \frac{d^{2} x}{{dt}^{2}}] But \frac{d^{2} x}{{dt}^{2}} = a_{x} = 0 . Hence a_{y} = - \frac{3}{2} {(\frac{dx}{dt})}^{2} = - \frac{3}{2} u^{2} x = - \frac{3}{2} \times (3)^{2} = - \frac{27}{2} {ms}^{- 2} Range, R = \frac{2 u_{x} u_{y}}{a_{y}} = \frac{2 \times 3 \times 36}{27 / 2} = 16 m Alternatively: We have y = 12 x - \frac{3}{4} x^{2} . When projectile again comes to ground, y = 0 and x = R . 0 = 12 R - \frac{3}{4} R^{2} \Rightarrow R = 16 m$

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