# Projectile Strikes a Wall

A particle $P$ is thrown horizontally from a tower of height $H$. After some time, it hits a wall and falls on the ground with an angle $\theta$ with the horizontal. Given that the distance between the tower and the wall is $D$ and the distance between the particle and the wall is $d$, what is the height of the tower $H$?

Honestly, I don’t know what should I do first to tackle this problem, since there are quite many different path that can lead to the same answer. So, the strategy I’d follow is backtracking. First, we note down some simple equations that might be useful afterwards. For example, the relationship between the horizontal partial speed and vertical partial speed.

\begin{aligned} \displaystyle \tan \theta &= \frac{v_y}{v_x} \\ v_y &= v_x \tan \theta \end{aligned}

The backtrack strategy is simple: We first arrive at the final equation, and continue solving for the missing variable until we break everything into their simplest form. In this case, $\theta, D, d$.

Solve for $H$,

\begin{aligned} \displaystyle {v_y}^2 - {u_y}^2 &= 2gH \\ H &= \frac{{v_y}^2}{2g} \end{aligned}

Now, we know that we have to solve for $v_y$.

\begin{aligned} \displaystyle v_y &= gt \end{aligned}

Then, we solve for $t$. We know that particle $P$ has traveled $(D+d)$ horizontal distance.

\begin{aligned} \displaystyle v_x &= \frac{D+d}{t} \\ t &= \bigg (\frac{D+d}{v_y}\bigg ) \tan \theta \end{aligned}

Substitute $t$ into the previous equation.

\begin{aligned} \displaystyle v_y &= g\bigg (\frac{D+d}{v_y}\bigg ) \tan \theta \\ {v_y}^2 &= (D+d)g\tan \theta \end{aligned}

Finally, we substitute ${v_2}^2$ into the first equation.

\begin{aligned} \displaystyle H &= \bigg(\frac{(D+d)g}{2g}\bigg) \tan \theta \\ &= \bigg(\frac{D+d}{2}\bigg) \tan \theta \end{aligned}