Speed

Oct 2012
1,314
21
USA
5656.jpg - Hint on beginning setup?
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
calculate the time the shell hit's the ground.

The maximum height will occur, assuming the initial and final heights are the same, at 1/2 the total flight time.

You can then calculate the max height in terms of the initial speed and determine the initial speed necessary to reach 542m.
 
Oct 2012
1,314
21
USA
\(\displaystyle y(t) = v_{0}\sin\theta(t) - \dfrac{1}{2}g(t)^{2} \)

\(\displaystyle (t)(v_{0}\sin\theta-\dfrac{1}{2}g(t)^{2}) = 0\)

\(\displaystyle t = 0\)

\(\displaystyle t = \dfrac{2v_{0}\sin\theta}{g}\)

\(\displaystyle x(\dfrac{2v_{0}\sin\theta}{g}) = v_{0}\cos\theta( \dfrac{2v_{0}\sin\theta}{g})\)

\(\displaystyle v_{0}\cos(30)( \dfrac{2v_{0}\sin(30)}{9.8}) = 542\)

\(\displaystyle v_{0}(\dfrac{\sqrt{3}}{2})( \dfrac{2v_{0}(\dfrac{1}{2})}{9.8} = 542\)

\(\displaystyle v_{0}^{2} = (\dfrac{\sqrt{2}}{\sqrt{3}})(\dfrac{1}{.50})(9.8)(\dfrac{1}{2})(542)\)

\(\displaystyle v_{0}^{2} = 6133.30738\) - ??

\(\displaystyle v_{0} = \sqrt{6133.30738} = 78.31543513 = 78.3\) - ??
 
Last edited:

romsek

MHF Helper
Nov 2013
6,836
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California
542 is the max height. Not the ground range. You had it right until you started messing with x(t).

You also seem to have ignored me about the max height being half the total flight time.
 
Oct 2012
1,314
21
USA
Max height would involve y'(t)

\(\displaystyle y'(t) = v_{0}\sin(30) - 9.8t = 0\) - ?? two variables
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Max height would involve y'(t)

\(\displaystyle y'(t) = v_{0}\sin(30) - 9.8t = 0\) - ?? two variables
So $t=\dfrac{v_0\sin(30)}{g}$

Substitute that value for $t$ into the $y(t)$ equation, then solve for $v_0$.
 
Oct 2012
1,314
21
USA
\(\displaystyle y(t) = v_{0}\sin(30)(t) - \dfrac{1}{2}g(t)^{2} \)

\(\displaystyle y(t) = v_{0}(\dfrac{1}{2})(t) - \dfrac{1}{2}(9.8)(t)^{2} \)

\(\displaystyle y(t) = v_{0}(\dfrac{1}{2})(t) - 4.9(t)^{2} \)

\(\displaystyle y'(t) = v_{0}(\dfrac{1}{2}) - 9.8t \)

\(\displaystyle v_{0}\dfrac{1}{2}- 9.8t = 0 \)

\(\displaystyle v_{0}\dfrac{1}{2} = 9.8(t) \)

\(\displaystyle t = \dfrac{v_{0}(\dfrac{1}{2})}{9.8}\)

\(\displaystyle y(t) = v_{0}\sin(30)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(g)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8})^{2} \)

\(\displaystyle y(t) = v_{0}\dfrac{1}{2}(\dfrac{v_{0}(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(9.8)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8})^{2} \)

\(\displaystyle 2 = (\dfrac{(v_{0})^{2}(\dfrac{1}{2})}{9.8}) - (9.8)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8})^{2} \)

\(\displaystyle 2 = (\dfrac{(v_{0})^{2}(\dfrac{1}{2})}{9.8}) - (9.8)(\dfrac{(v_{0})^{2}}{\dfrac{2401}{25}}) \)

\(\displaystyle 2 = (\dfrac{(v_{0})^{2}(\dfrac{1}{2})}{9.8}) - (9.8)(\dfrac{(v_{0})^{2}}{\dfrac{2401}{25}}) \)

\(\displaystyle \dfrac{2}{(v_{0})^{2}} = (\dfrac{(\dfrac{1}{2})}{9.8}) - (9.8)(\dfrac{1}{\dfrac{2401}{25}}) \)

\(\displaystyle \dfrac{1}{(v_{0})^{2}} = (\dfrac{(\dfrac{1}{2})}{9.8})(\dfrac{1}{2}) - (9.8)(\dfrac{1}{\dfrac{2401}{25}}) (\dfrac{1}{2}) \)

\(\displaystyle \dfrac{1}{(v_{0})^{2}} = -\dfrac{5}{196} \)

\(\displaystyle (v_{0})^{2} = -\dfrac{196}{5}\)

\(\displaystyle (v_{0}) = \sqrt{-\dfrac{196}{5}}\) ??? - square roots cannot have a negative number inside
 
Last edited:
Oct 2012
1,314
21
USA
Alternate method: ?

\(\displaystyle y(t) = v_{0}\sin(30)(t) - \dfrac{1}{2}g(t)^{2} \)

\(\displaystyle y(t) = v_{0}(\dfrac{1}{2})(t) - \dfrac{1}{2}(9.8)(t)^{2} \)

\(\displaystyle y(t) = v_{0}(\dfrac{1}{2})(t) - 4.9(t)^{2} \)

\(\displaystyle y'(t) = v_{0}(\dfrac{1}{2}) - 9.8t \)

\(\displaystyle v_{0}\dfrac{1}{2}- 9.8t = 0 \)

\(\displaystyle v_{0}\dfrac{1}{2} = 9.8(t) \)

\(\displaystyle t = \dfrac{v_{0}(\dfrac{1}{2})}{9.8}\)

\(\displaystyle y(t) = v_{0}\sin(30)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(g)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8})^{2} \)

\(\displaystyle y(t) = v_{0}\dfrac{1}{2}(\dfrac{v_{0}(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(9.8)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8})^{2} \)

\(\displaystyle y(t) = (v_{0})^{2}\dfrac{1}{2}(\dfrac{(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(9.8)(\dfrac{v_{0}(\dfrac{1}{2})}{9.8})^{2} \)

\(\displaystyle y(t) =(v_{0})^{2} \dfrac{1}{2}(\dfrac{(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(9.8)((\dfrac{v_{0})^{2}(\dfrac{1}{4})}{\dfrac{2401}{25}}\)

\(\displaystyle \dfrac{1}{(v_{0})^{2}} = \dfrac{1}{2}(\dfrac{(\dfrac{1}{2})}{9.8}) - \dfrac{1}{2}(9.8)((\dfrac{(\dfrac{1}{4})}{\dfrac{2401}{25}}\)

\(\displaystyle \dfrac{1}{(v_{0})^{2}} = \dfrac{5}{196} - \dfrac{5}{392}\)

\(\displaystyle \dfrac{1}{(v_{0})^{2}} = \dfrac{5}{396}\)

\(\displaystyle (v_{0})^{2} = \dfrac{396}{5}\)

\(\displaystyle v_{0} = \sqrt{396}{5} = \dfrac{6\sqrt{55}}{5} = 8.899438185\) ??
 
Last edited:

romsek

MHF Helper
Nov 2013
6,836
3,079
California
you are making this so flippin hard...

$y(t) = v_0 \sin(\theta) t - \dfrac {g t^2}{2}$

at $t_f$, the shell has hit the ground and $y(t)=0, t>0$

$v_0 \sin(\theta) t_f = \dfrac {g t_f^2}{2}$

$v_0 \sin(\theta) = \dfrac {g t_f}{2}$

$t_f = \dfrac {2 v_0 \sin(\theta)}{g}$

Now the maximum height occurs at $\dfrac {t_f}{2}= \dfrac {v_0 \sin(\theta)}{g}$

$y_{max}=y\left(\dfrac {t_f}{2}\right) = v_0 \sin(\theta) \left(\dfrac {v_0 \sin(\theta)}{g}\right) - \dfrac {g}{2} \left( \dfrac {v_0 \sin(\theta)}{g}\right)^2$

$y_{max}=\dfrac{v_0^2 \sin^2(\theta)}{g} -\dfrac{v_0^2 \sin^2(\theta)}{2g}= \dfrac{v_0^2 \sin^2(\theta)}{2g}$

Dumping values in

$\sin^2(30^\circ)=\dfrac 1 4$

$g=9.8$

$542 = \dfrac 1 4 \dfrac {v_0^2}{19.6}$

$542 \times 4 \times 19.6 = 42492.8 = v_0^2$

$v_0 = \sqrt{42492.8} \approx 206.14$
 
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