# Speed

#### Jason76

- Hint on beginning setup?

#### romsek

MHF Helper
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.

#### Jason76

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

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

$$\displaystyle t = 0$$

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

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

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

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

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

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

$$\displaystyle v_{0} = \sqrt{6133.30738} = 78.31543513 = 78.3$$ - ??

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#### romsek

MHF Helper
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.

#### Jason76

Max height would involve y'(t)

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

#### skeeter

MHF Helper
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 $v_0$.

#### Jason76

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\displaystyle (v_{0}) = \sqrt{-\dfrac{196}{5}}$$ ??? - square roots cannot have a negative number inside

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#### Jason76

Alternate method: ?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$\displaystyle v_{0} = \sqrt{396}{5} = \dfrac{6\sqrt{55}}{5} = 8.899438185$$ ??

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#### romsek

MHF Helper
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

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

$g=9.8$

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

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

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

1 person

#### Jason76

What happened to 542?