Velocity of a Projection In Projectile Motion

[seallen]

Homework Statement

For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substantiate your answer with mathematical support.

Homework Equations

My teacher gave us the information that u=10 m/sec, however I don't see how she arrived at that conclusion, also where did the angles of projection come from?

The Attempt at a Solution

if u = 10 m/sec , and angles of projection are 30o, 45o and 50o then,
the horizontal distances traveled are
R1 = 10^2 x sin 60 / 10 = 5 square root of 3 m
R2= 10^2 x sin 90 / 10 = 10 m
R3 = 10^2 x sin 100 / 10 = 10 x 0.9848 = 9.848 m
thus, you see that the distance is maximum for 45o

 

[bossman27]

She probably wants you to derive an expression horizontal distance with an unknown θ, and then use your knowledge of trig functions to explain why the distance is maximized at θ=45°. I imagine that she gave you an initial velocity to work with simply so that you would have one less variable to be confused by, and obviously the θ you come up with will be independent of what the initial velocity is (so long as it's greater than 0).

 

[Spinnor]

It looks like you have the right expression for the range of the projectile, R, R(theta, velocity) = you know what it is. Now take the derivative of R with respect to theta (velocity held fixed) and set that = to zero and solve for theta.