Question

How to find the total horizontal displacement for a projectile and time in the air given angle and velocity?

Answer 1

You use the angle to find Vx and Voy. It should go like this:
\[V _{0y} = V _{given}\sin \theta\]\[V _{x} = v _{given} \cos \theta\]Now find t using the equation\[v _{f} = v _{0}+g t\]After that use one of the following two equations:\[\Delta x = v _{0}t + \frac{ 1 }{ 2 } g t ^{2}\]with\[v _{f} = 0\]
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Use t to find x (distance).\[\Delta x = \frac{ 1 }{ 2 }\left( v _{0}+v _{f} \right)t\]There is another equation but it's harder, because it creates a quadratic that you must solve:\[\Delta x = v_{0}t + \frac{ 1 }{ 2 }g t ^{2}\]

Answer 2

Then multiply x by 2.

Answer 3

*typo: there are not two equations, just one. and the second equation on the bottom isn't harder, I thought you were finding time at first xD

Answer 4

How do you find the maximum height? @kittiwitti1

Answer 5

Just replace the delta x with delta y, Vo with Voy, and Vf with Vfy; however you don't multiply by 2 because the maximum height is when the projectile is halfway through its parabolic path.

Answer 6

Here, I'll just type it out for you.

Also, there was one typo in the equations of the first part.
It's not g but a (acceleration), only for y values does gravity (9.8 m/s/s downward force) apply.
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You don't need the Vx for this one, just as I accidentally typed the Voy for the first part's trig equations. But you can find them together in the first part in order to save time/space, or for convenience if you so desire.
\[V _{oy} = V _{given}\cos \theta \]Now input this into the equation to find t (Vfy = 0 so that cancels out):
\[v _{fy} = v _{0y} + g t\]Now that you've fond t, use that to find y-distance. You choose which formula from the two below that you want to use. Neither of them are truly easier than the other, it's just preference really.\[\Delta y = \frac{ 1 }{ 2 }\left( v _{oy} + v_{fy} \right)t\]\[\Delta y = v _{0y}t + \frac{ 1 }{ 2 }g t ^{2}\]Remember, Vfy cancels out to 0, NOT Voy. That's if it starts at rest, which gives it no vertical velocity (there could possibly be a horizontal velocity though, which would be Vx).