First slide

Projection Under uniform Acceleration

Question

The equation of motion of a projectile is : y = 12x- $\frac{3}{4} x^{2}$
The horizontal component of velocity is 3 ${ms}^{- 1}$ . Given that g = 10 ${ms}^{- 2}$ , what is the range of the projectile?

Moderate

A

12.4 m
B

21.6 m
C

30.6 m
D

36.0 m

Solution

y = 12x - $\frac{3}{4} x^{2}$
$\frac{dy}{dt} = 12 \frac{dx}{dt} - \frac{3}{2} x \frac{dx}{dt}$
At x = 0, $\frac{dy}{dt} = 12 \frac{dx}{dt}$
If $θ$ be the angle of projection, then
$\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 12 = \tan θ$
Also, if u = initial velocity, then u $\cos θ$ = 3
Hence, $\tan θ \times \cos θ = 36 or u \sin θ = 36$
Range R = $\frac{u^{2} \sin 2 θ}{g} = \frac{2 u^{2} \sin θ \cos θ}{g}$
= $\frac{2 (u \sin θ) (u \cos θ)}{10} = \frac{2 \times 36 \times 3}{10} = 21.6 m$

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