If Roger is fired from the cannon with an angle of inclination θ of 60° and that he hits...

Question

If Roger is fired from the cannon with an angle of inclination θ of 60° and that he hits the ground 1/2 mile from the cannon. What, then, was Roger's initial speed?

Answer 1

We have been given the range here. So we can go forward with: $$R = \frac{v^2Sin2θ}{g}$$

where,

R = Range in Km, here given 1/2 miles which is = 0.804672 Km = 804.672 m

v = Initial Velocity which we need to find.

θ = $60^0$

g = 9.80665 $m/s^2$

Now the v value can be easily calculated which is the initial velocity.

Answer 2

suppose the acceleration of gravity is $g$, the time for hiting the ground is $t$, and the initial speed is $v$, then we have the velocity components are $$v_x=v\cos(60^\circ)=\frac{v}{2}\\v_y=v\sin(60^\circ)=\frac{\sqrt{3}v}{2}$$ According to the formulas of velocity and accelerration, we have the equations: $$v_xt=\frac{1}{2}\\v_y=g\frac{t}{2}$$ then $$vt=1\\\sqrt{3}v=gt$$ $\Rightarrow$ $$t=\frac{1}{v}\\\sqrt{3}v^2=g$$ so $$v=\sqrt{\frac{g}{\sqrt{3}}}$$