Projectile Strikes a Wall

christopherboo66, ,

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?

Screenshot 2015-01-25 at 18.33.13

Problem adapted from Brilliant.org

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}

Projectile Strikes a Wall

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