The question in the calculus class is If the initial velocity of a projectile is $80$ ft/s with an angle of elevation of $55^\circ$, at what height will it strike a building that is $178$ ft away from the launching point?
The answer is given as $13.4$ ft. I understand my projectile motion equations, but it's not clear to me which equations to use to produce this solution. Any enlightenment would be greatly appreciated. It's not clear to me why this is even given in a calculus class. I thought I was signing up for a calc class and instead I'm doing physics problems.
$v_0=80\,\mathrm{ft\,s}^{-1}$, $\theta=55^\circ$, $d=178\,\mathrm{ft}$, $g\approx 32\,\mathrm{ft\,s}^{-2}$. The height $y(x)$ of the projectile at distance $x$ is given by $$ y(x)=y_0+x\tan \theta -{\frac {gx^{2}}{2(v_0\cos \theta )^{2}}} $$ We have $y_0=0$, and at $x=d$ we find the height $$ h=y(d)=d\tan \theta -{\frac {gd^{2}}{2(v_0\cos \theta )^{2}}}=13.4\,\mathrm{ft} $$