Multiple choice

The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, find the horizontal range of the projectile.

  1. R + H

  2. R + 2H

  3. R

  4. R + (H)/2

  5. 2R
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B Correct answer
Explanation

R = (2 u2 Sin θ Cos θ)/gH = (u Sin θ)2/ 2gT = (2 u Sin θ)/g(Time of flight = t) Now when wind blows in horizontal direction R (new) = u Cos θ*t + 1/2 (g/2) t2 (Applying equation of motion) R (new) = (2 u2 Sin θ Cos θ)/g + (g/4) (4 u2 Sin2θ)/g2 Solving we get R (new) = R + (u2 Sin2θ)/g R (new) = R + 2H (Answer)

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