Question

Find the initial speed which a projectile must be subjected to give it a maximum horizontal range of 490m. Assume the acceleration due to gravity as g=10m/s

Answer 1

$$x(t)=v_{0x}t=v_0\cos \alpha t$$

$$y(t)=v_{0y}t-\dfrac{gt^2}{2}=v_0\sin \alpha t-\dfrac{gt^2}{2}$$

$$t_{fligt}=\dfrac{2v_{0y}}{g}=\dfrac{2v_{0}\sin \alpha}{g}$$

Then

$$x_{flight}=v_0\cos \alpha(\dfrac{2v_{0}\sin \alpha}{g})=\dfrac{v_{0}^2\sin(2 \alpha)}{g}$$

A maximum horizontal range will be at $\alpha =45\degree.$ We have

$$v_0=\sqrt{x_{max}g}$$

$$v_0=\sqrt{490m(10m/s^2)}=70m/s$$