First slide

Projection Under uniform Acceleration

Question

A projectile is projected with initial velocity $(6 \hat{i} + 8 \hat{j}) m / s .$ If g = 10 m/s², then what is the horizontal range of the projectile?

Easy

A

8.8 m
B

13.5 m
C

9.6 m
D

11.2 m

Solution

Initial velocity is $(6 \hat{i} + 8 \hat{j}) m / s$ (given).
Magnitude of velocity of projection is $u = \sqrt{u_{x}^{2} + u_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$
Angle of projection $tanθ = \frac{u_{y}}{u_{x}} = \frac{8}{6}$
$\Rightarrow sinθ = \frac{4}{5} and cosθ = \frac{3}{5}$
Now the horizontal range is, $R = \frac{u^{2} \sin 2 θ}{g} = \frac{u^{2} 2 \sin θ \cos θ}{g}$
$= \frac{{(10)}^{2} \times 2 \times (4 / 5) \times (3 / 5)}{10} = 9 .6 m$

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