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Grade: 11

                        

A ball is thrown eastwards across a level ground.A wind blows horizontally tothe 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.find the angle (with horizontal ) at which the ball should be projected for maximum range? pls as quickly!!!!!!

4 years ago

Answers : (4)

Vikas TU
12133 Points
							
This is simalr to the motion of projectile as whatevr method we always apply to it.
The thing that is different is only that the acceleration in horizontal direcn is not zero or constant.
it is also g just like in y – direction.
4 years ago
Prsahant Parashar
46 Points
							I thi
						
4 years ago
Prsahant Parashar
46 Points
							
I think the angle is of 45 as it provides the maxmimum range 
and the wind is also blowing in the same direction and it gives force of egual magnitude to all the angles so for max range angle should be of 45
4 years ago
Ransom
13 Points
							here the acceleration is g both horizontally (eastward) and vertically (downward) 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° hope this helps
						
3 years ago
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