Question

Derive an expression for the range of a projectile. What angle x from the horizontal gives the maximum range?

Answer 1

45' from the ground gives a projectile a max range horizontaly. . .

Answer 2

I will call alpha the initial angle (not x) \[x(t)= v_xt, y(t)=v_yt-\frac{1}{2} g t^2 \] because the only force is gravity and is acting downward (we are in a uniform gravitational field). We want y=0, then t=0 (starting position) or \[t=\frac{2v_y}{g}\] The corresponding x is therefore \[x=\frac{2v_x v_y}{g}\] Now as I said let alpha be the angle of the velocity vector with the ground, then we have \[v_x=v\cos\alpha, v_y=v\sin \alpha \] And finally \[x=\frac{2v^2\cos\alpha\sin\alpha}{g}=\frac{v^2\sin 2\alpha}{g}\] The maximum x is then for alpha=45° and is equal to \[x_{max}=\frac{v^2}{g}\]

Answer 3

Now answer the same question with a friction force opposing the velocity.

Answer 4

thnx for teh reply!!!!