Question

A ball is thrown eastward across level ground. A wind blows horizontally to the east, and assume that
the effect of wind is to provide a constant force towards the east, equal in magnitude to the weight of
the ball. The angle 'thetha' (with horizontal east) at which the ball should be projected so that it travels
maximum horizontal distance is
(A) 45°
(B) 37°
(C) 53°
(4) 67.5°​

Answer 1

Answer:

67.5°

Explanation:

The acceleration of the body is g.

Time of flight = t = 2usinθ / g

horizontally s = ut + (0.5)at²

s = ucosθ(2usinθ)/g + (0.5)g[2usinθ / g]²

s = (1/g)[u²sin2θ + 2u²sin²θ]

for maximum s => ds/dθ = 0

=> (u²/g)[cos2θ.2 + 4sinθcosθ] = 0

=> [cos2θ + sin2θ] = 0 or tan2θ = -1

=> θ = 135/2 = 67.5°

also the max range R = 2.555(u²/g)

Ans: angle θ the ball should be thrown so that it travels the

maximum horizontal distance = 67.5°

Answer 2

Answer:

The acceleration of the body is g.

Time of flight = t = 2usinθ / g

horizontally s = ut + (0.5)at²

s = ucosθ(2usinθ)/g + (0.5)g[2usinθ / g]²

s = (1/g)[u²sin2θ + 2u²sin²θ]

for maximum s => ds/dθ = 0

=> (u²/g)[cos2θ.2 + 4sinθcosθ] = 0

=> [cos2θ + sin2θ] = 0 or tan2θ = -1

=> θ = 135/2 = 67.5