Projectile motion maximum range

  • #1
A plane is inclined at an angle A to the horizontal. A particle is projected up the plane with a velocity u at an angle B to the plane. the plane of projection is vertical and contains the the line of greatest slope

Prove that the range is a maximum when [tex]\displaystyle{B=\frac{1}{2}[(\frac{\pi}{2})-A]}[/tex]

I found the time of flight to be [tex]\displaystyle{\frac{2uSinB}{gCosA}}[/tex] and then tried to find [tex]S_{x}[/tex] at this time. It was pretty long and im not sure how to prove the range bit. Is it something to do with the maximum value of Sine being 1?

Thank you
 

Answers and Replies

  • #2
I think you're on the right track with the values of sine. Since you can break up the motion of a projectile into x and y components by taking the sine and cosine of the same angle, the maximum range will be when the sum of both x and y components are their largest. The larger the y value, the more flight time there is; the larger the x value is, the more range it gets per unit time. For example, at 90 degrees to the horizontal: sine is 1 but cosine is zero, so the sum of sin(90) and cos(90) is 1 (shooting straight up, all y component, no x component, not maximum range). At 85 degrees to the horizontal: sin(85) = .996 and cos(85) = .087, sum is 1.083, small x componenet, large y componenet, not maximum range. At 45 degrees: sin(45) is .707 and cos(45) is .707, and their sum is 1.414 (maximum range).

This is just a guess/shot in the dark but I hope it helps.
 

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