Projectile motion maximum range

  • #1
No Name Required
29
0
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 I am 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
hologramANDY
16
0
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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