I have done this 5 different ways. And all wrong. I need help. A projectile is l

Question

I have done this 5 different ways. And all wrong. I need help.

A projectile is launched upward at an angle. What should thisangle be with respect to the horizontal in order for the projectileto travel the farthest distance? Explain why.

Explanation / Answer

Assuming no air resistance, 45 degrees. Going up and coming back down, we use the projectile motionequation, 0m = (vsin)*t - (1/2)(g)(t2) => t =2vsin/g Now, that gives us the time using the vertical motion. The horizontal motion is given by d = vcos*t substituting t in, we find d = 2v2cossin /g take the derivative of d with respect to , we find d' = 2v2/g (cos2 -sin2) d' = (2v2/g) (cos(2) to find the max, set d' = 0. Then divide out the constants. 0 = cos(2) 2 = 90 deg = 45

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