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?
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what is the acceleration due to gravity? what simplifying assumptions do we make? and more importantly, what have you tried? – Guy Mar 19 '14 at 08:21
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acceleration due to gravity is 9.8m/s^2. I'm only able to get as far as using y(t)=-(1/2)gt^2 + (vsinθ)t + h, where g and theta and h are known but not sure what to use for t. @Sabyasachi – user136446 Mar 19 '14 at 08:35
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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.
Kailas
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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 $$v=\sqrt{\frac{g}{\sqrt{3}}}$$
Martial
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you give that the angle of inclination θ is $60^\circ$, so $v_x=v\cos(60^\circ)=v\sin(30^\circ)$ – Martial Mar 19 '14 at 08:49
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Sorry, meant to ask why is it 30 before you edited it. Thanks though. – user136446 Mar 19 '14 at 08:54
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It's just the relationship beween the right-angle sides and the hypotenuse in a right-angled triangle. I wrote $30$ because $30=90-60$, it's not a problem if you want to write 60 – Martial Mar 19 '14 at 09:02