Hi! Is it possible to determine the angle (2 dimensional) that a projectile was launched given the gravity, initial velocity (but again, no angle given) and the final horizontal and vertical displacements? I tried solving for when the vertical velocity was zero (when it reached the peak), but I didn't make it far. Thanks in advance.

Question

matt101 · Answer

Break the velocity down into horizontal and vertical components and find equations for them:

y = vsinθt - (1/2)gt^2
x = vcosθt

If you're given both the horizontal and vertical displacements of a particular point, you know they have the same value for t. Rearrange the second equation to solve for t, and sub that into the first equation:

t = x/(vcosθ)
y = vsinθ(x/(vcosθ)) - (1/2)g(x/(vcosθ))^2

If you're given values for every other variable, you only have to solve for θ.

anonymous · Answer

but won't that give me 2 possible values for t?

anonymous · Answer

oh haha i see never mind. thank you very much!