In a projectile motion problem, how do you solve for maximum height when given only horizontal distance and angle of launch?

Answer 1

The short answer is to set up the equations of motion and solve them.

Let

α be the angle of launch,

v be the velocity of launch (unknown),

g be gravitational acceleration,

x(t) be be the horizontal distance after time t,

y(t) be the vertical distance after time t.

The trajectory is a parabola resulting from a combination of two motions:

A straight line due to inertia, which the projectile would follow in the absence of gravity, and

downward fall, which the projectile would suffer in the absence of launch velocity.

The x distance is not affected by gravity, so

x(t) = vtcos(α)

The vertical distance is a combination of upward inertial flight and downward fall

y(t) = vtsin(α) - gt²/2

The projectile will fall back to earth at some time t₁ satisfying

y(t₁) = 0

So you have two equations

x(t₁) = vt₁cos(α)

0 = vt₁sin(α) - gt₁²/2

You know x(t₁), α, g, and have only the two unknowns v and t₁.

You can solve the two equations for v and t₁, and then get the answer to your question by evaluating

y(t₁/2)