Question

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 horizontal
motion of projectle, giving it a constant horizontal
acceleration equal to g. Under the same conditions
of projection, the new range will be
(g = acceleration due to gravity)
(1) R+H
(2) R+2H
(4) R+ 4H
(3) R+ 3H/2
​

Answer 1

Answer:

Explanation:

=> Here, the horizontal range of a projectile is R and the  maximum height attained by it is H.

Thus, maximum height H = μ² sin² θ / 2g

2H = μ² sin² θ / g

R = 2μ² sin θ * cos θ / g

=> When a strong wind  begins to blow :

time of flight, T = 2μsinθ / g

Sy = μy*t + 1/2a_y*t²

0 = μy*T + 1/2a_y*T²

0 = T (μy - 1/2gT)

T = 2μy / g

=> Now, displacement along x - axis:

Su = μₓT + 1/2 *aₓ * T²

= μ cos θ * 2μsinθ/g + 1/2*g*(2μsinθ/2)²

Range = R + 2/g* μ²sin²θ [μ²sin²θ/g = 2H ]

           = R + 2 * 2H

           = R + 4H