A daredevil is shot out of a cannon

[karush]

$\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*}

 

[HallsofIvy]

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 \(\displaystyle \left<\frac{25\sqrt{2}}{2}, \frac{25\sqrt{2}}{2}\right>\) (that "\(\displaystyle \frac{\sqrt{2}}{2}\)" is your "[FONT=MathJax_Main]cos[/FONT][FONT=MathJax_Main]45[/FONT][FONT=MathJax_Main]∘[/FONT]) so the velocity vector is \(\displaystyle \left<\frac{25\sqrt{2}}{2}, \frac{25\sqrt{2}}{2}- gt\right>\) and, taking the initial position to be (0, 0), the position vector is \(\displaystyle \left<\frac{25\sqrt{2}}{2}t, \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2\right>\). 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 \(\displaystyle \left<\frac{25\sqrt{2}}{2}t, \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2\right>= <50, -h>\). That is the same as the two equations \(\displaystyle \frac{25\sqrt{2}}{2}t= 50\) and \(\displaystyle \frac{25\sqrt{2}}{2}t- \frac{1}{2}gt^2= -h\).

Solve the first equation for t then put that value of t into the second equation to find h.

 

[skeeter]

$\Delta x = v_0 \cos{\theta_0} \cdot t \implies t = \dfrac{\Delta x}{v_0 \cos{\theta_0}}$

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

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 $\Delta y$.