First slide

Projectile motion

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?

Moderate

A

18 m
B

16 m
C

12 m
D

21.6 m

Solution

$\begin{matrix} Given y = 12 x - \frac{3}{4} x^{2}, u_{x} = 3 {ms}^{- 1} \\ v_{y} = \frac{d y}{d t} = 12 \frac{d x}{d t} - \frac{3}{2} x \frac{d x}{d t} \\ At x = 0, v_{y} = u_{y} = 12 \frac{d x}{d t} = 12 u_{x} = 12 \times 3 = 36 {ms}^{- 1} \\ a_{y} = \frac{d}{d t} (\frac{d y}{d t}) = 12 \frac{d^{2} x}{d t^{2}} - \frac{3}{2} [{(\frac{d x}{d t})}^{2} + x \frac{d^{2} x}{d t^{2}}] \end{matrix}$
$But \frac{d^{2} x}{d t^{2}} = a_{x} = 0 . Hence$
$a_{y} = - \frac{3}{2} {(\frac{d x}{d t})}^{2} = - \frac{3}{2} u_{x}^{2} = - \frac{3}{2} \times (3)^{2} = - \frac{27}{2} {ms}^{- 2}$
$Range, R = \frac{2 u_{x} u_{y}}{a_{y}} = \frac{2 \times 3 \times 36}{27 / 2} = 16 m$
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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