# Mechanics: calculating launch angle of projectile

• roam
In summary, the conversation revolved around finding the launch angle of a ball using the relationship ##\theta=\arctan(\frac{v_{y}}{v_{x}})##. The horizontal velocity was calculated to be approximately 21 m/s, but the vertical velocity was not given. The problem was approached by considering the vertical motion separately, with the start position at y=0 and the final position at y=10m. The relevant equations, ##v^2_y = v^2_{0y} - 2 g(y-y_0)## and ##v^2_y = v_{0y}^2 -2a(y-y_0)##, were discussed but could not
kuruman said:
You can invent your own equation in terms of the given quantities. Knowing the time of flight makes the task easy. You know that the maximum height occurs at half the time of flight and that the vertical velocity is zero when that happens. Then you can write two equations expressing these ideas:
$$h_{\text{max}}=v_{\text{0y}}\frac{t_{\!f}}{2}-\frac{1}{2}g\left(\frac{t_{\!f}}{2}\right)^2~;~~0=v_{\text{0y}}^2-2gh_{\text{max}}.$$The task before you is to use the second equation to substitute ##v_{\text{0y}}## into the first equation and solve the ensuing quadratic.
You do need to be careful if ##t_f## is not when it returns to the ground.

PeroK said:
You do need to be careful if ##t_f## is not when it returns to the ground.
Oops! I lost myself somewhere.

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