The horizontal range and maximum height attained by a projectile are R and H- respectively- If a constant horizontal acceleration a-g4 is imparted to the projectile due to wind- then its horizontal range and maximum height will be

As we know, H=u^2 sin^2⁡θ2g,R=u^2 sin⁡2θg No change is y direction, Hence Maximum height will remain the same. For Range, Given, horizontal acceleration a=g4 R ...

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Standard XII

Physics

Projectile Time, Height and Range

The horizonta...

Question

The horizontal range and maximum height attained by a projectile are $R$ and $H$ , respectively. If a constant horizontal acceleration $a = \frac{g}{4}$ is imparted to the projectile due to wind, then its horizontal range and maximum height will be

(R + H), \frac{H}{2}

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(R + H), 2 H

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(R + 2 H), H

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(R + H), H

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Solution

The correct option is D $(R + H), H$
$H = \frac{u^{2} {sin}^{2} θ}{2 g}, R = \frac{u^{2} sin 2 θ}{g}$ Maximum height will be same because acceleration $a = g$ is in horizontal direction.
$R^{'} = u cos θ T + \frac{1}{2} a T^{2}$
Since time of flight, $T = \frac{2 u sin θ}{g}$
$∴ R^{'} = u cos θ (\frac{2 u sin θ}{g}) + \frac{1}{2} \frac{g}{4} {(\frac{2 u sin θ}{g})}^{2}$
$∴ R^{'} = R + \frac{1}{2} \frac{g}{4} {(\frac{2 u sin θ}{g})}^{2} = R + H$

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The horizontal range and maximum height attained by a projectile are R and H, respectively. If a constant horizontal acceleration a=g4 is imparted to the projectile due to wind, then its horizontal range and maximum height will be

The correct option is D (R+H),HH=u2sin2θ2g,R=u2sin2θg Maximum height will be same because acceleration a=g is in horizontal direction. R′=ucosθT+12aT2 Since time of flight, T=2usinθg ∴R′=ucosθ(2usinθg)+12g4(2usinθg)2 ∴R′=R+12g4(2usinθg)2=R+H

The horizontal range and maximum height attained by a projectile are $R$ and $H$ , respectively. If a constant horizontal acceleration $a = \frac{g}{4}$ is imparted to the projectile due to wind, then its horizontal range and maximum height will be