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projectile


Point shoot by initial velocity $v_0=600m/s$ along the direction with the angle about horizontal plane $\alpha =45{}^{\circ}$. Solve for the maximum height and the range(assuming $g\approx 10m\left/s^2\right.$,without air resistance).

$v_x=v_0\text{Cos}[\alpha ]$

$S=v_xt=v_0\text{Cos}[\alpha ]t=600 \frac{\sqrt{2}}{2}t$

$h=v_y(t/2)-\frac{1}{2}g (t/2)^2$

$v_y=v_0\text{Sin}[\alpha ]=g t/2$

$t=2\left.v_y\right/g=2\times 600\left.\frac{\sqrt{2}}{2}\right/10=60\sqrt{2}$

$S=300\sqrt{2}60\sqrt{2}=36000$

$h=\left(600\frac{\sqrt{2}}{2}\left(30\sqrt{2}\right)-\frac{1}{2}10\left(30\sqrt{2}\right)^2\right)$=9000

So Range=36000m, Height=9000m


Any problems?

My steps may be not so clear, how to improve the writing?

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The numbers are correct. I like words between formulas. –  André Nicolas Jun 27 '13 at 3:32
    
Why the $t/2$ in the equation for $h$ instead of $t$? $v_y$ (assuming it is the original vertical velocity) should be $v_0 \sin [\alpha]$ –  Ross Millikan Jun 27 '13 at 3:40
    
@RossMillikan t is the whole time, t1=the point fly to height and then t2=the point down to the earth, t1=t2=t/2. yes, something wrong in the writing of $v_y$ they only equal in the absolute value. if $v_y$ is scalar, then no problem. –  HyperGroups Jun 27 '13 at 3:46
    
That is why André Nicolas likes words. Usually $h$ is $h(t)$ for any given $t$. As he says, the final result is fine. –  Ross Millikan Jun 27 '13 at 3:51

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