Find the speed and direction of a particle which , when projected form a point 15m above the horizontal ground, just clears the top of a wall 26.25 m and 30m away.

Do not know how to figure out, as there are 2 variables, namely V and tan(theta)

that is using Cartesian motion equation.

actually, there are an infinite number of solutions depending on how one interprets

... just clears the top of a wall 26.25 m and 30m away.

I'll make the assumption that this point is the highest point of the particle's trajectory.

let \(\displaystyle v\) = initial velocity

\(\displaystyle \theta\) = launch angle relative to the horizontal

\(\displaystyle \Delta x = v \cos{\theta} \cdot t = 30\)

\(\displaystyle \Delta y = 26.25-15 = v \sin{\theta} \cdot t - \frac{1}{2}gt^2\)

since it just clears the wall, the y-component of velocity at that point is zero ...

\(\displaystyle 0 = (v\sin{\theta})^2 - 2g(26.25-15)\)

solve the system of equations.