First slide

Projection Under uniform Acceleration

Question

The equation of motion of projectile is
$y = 12 x - (3 / 4) x^{2}$
The horizontal component of velocity is 3 m/s. What is the range of projectile ? (g = 1.0 m / s²)

Moderate

A

12.4 m
B

21.6 m
C

30.6 m
D

36.0 m

Solution

Given that, $y = 12 x - (3 / 4) x^{2}$
$\begin{array}{l} ∴ \frac{dy}{dt} = 12 (\frac{dx}{dt}) - \frac{3}{2} x (\frac{dx}{dt}) \\ At x = 0, \frac{dy}{dt} = 12 (\frac{dx}{dt}) \end{array}$
If angle of projection be $θ$ , then
$\frac{(dy / dt)}{(dx / dt)} = \tan θ = 12$
If u is initial velocity of projection, then $ucos θ = 3$
Therefore, $\tan θ \times (ucos θ) = 36$
or $usin θ = 36$
Now, $R = \frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} \times 2 \sin θcos θ}{g}$
$= \frac{2 (usin θ) (ucos θ)}{g} = \frac{2 \times 36 \times 3}{10} = 21 \cdot 6 cm$

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