MHB A daredevil is shot out of a cannon

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A daredevil is launched from a cannon at a 45-degree angle with an initial speed of 25.0 m/s, aiming to land in a net positioned 50.0 m away. The initial velocity vector is calculated as <17.68, 17.68 - gt>, where g represents the acceleration due to gravity. To find the height of the net above the cannon, the position equations are set equal to the net's coordinates. By solving the horizontal distance equation for time and substituting it into the vertical position equation, the height can be determined. The final calculation involves substituting known values into the derived formula for vertical displacement.
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$\textsf{ A daredevil is shot out of a cannon at $45^\circ$ to the horizontal with an initial speed of 25.0 m/s.}$
$\textsf{ A net is positioned a horizontal distance of 50.0 m from the canon.}\\$
$\textit{At what height above the cannon should the net be placed in order to catch the daredevil?}$
so far did this
\begin{align*}\displaystyle
v&=(25)\cdot\cos{45^\circ}=17.68
\end{align*}
 
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And "what you have done" makes no sense because you haven't said what "v" means! And it can't be "velocity" because the velocity here is a vector quantity.

Here is what I would do- this is "acceleration due to gravity so the acceleration vector is <0, -g>. The initial velocity vector is [math]\left<\frac{25\sqrt{2}}{2}, \frac{25\sqrt{2}}{2}\right>[/math] (that "[math]\frac{\sqrt{2}}{2}[/math]" is your "[FONT=MathJax_Main]cos[FONT=MathJax_Main]45[FONT=MathJax_Main]∘) so the velocity vector is [math]\left<\frac{25\sqrt{2}}{2}, \frac{25\sqrt{2}}{2}- gt\right>[/math] and, taking the initial position to be (0, 0), the position vector is [math]\left<\frac{25\sqrt{2}}{2}t, \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2\right>[/math]. If the initial height of the cannon is "h" meters above the net, and the net is 50 meters from the cannon, then the net's position is <50, -h> so we must have [math]\left<\frac{25\sqrt{2}}{2}t, \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2\right>= <50, -h>[/math]. That is the same as the two equations [math]\frac{25\sqrt{2}}{2}t= 50[/math] and [math]\frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2= -h[/math].

Solve the first equation for t then put that value of t into the second equation to find h.
 
$\Delta x = v_0 \cos{\theta_0} \cdot t \implies t = \dfrac{\Delta x}{v_0 \cos{\theta_0}}$

$\Delta y = v_0 \sin{\theta_0} \cdot t - \dfrac{1}{2}gt^2$

substitute the expression for $t$ derived from the first equation into the second ...

$\Delta y = \Delta x \cdot \tan{\theta_0} - \dfrac{g \cdot (\Delta x)^2}{2v_0^2 \cos^2{\theta_0}}$

you were given values for $\Delta x$, $v_0$, and $\theta_0$ ... calculate the value of $\Delta y$.
 
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