Quote:
Originally Posted by Freebase Dali
Word. Ok. So I have the basics. Now, when doing 2 dimensional kinematics, that trig comes into play, but I also have to remember some formulas. I can put those on my phone, so I'm not worried too much about it, since my phone is also my calculator, which is allowed.
So, how would you guys go about solving something like this (taken from a homework problem):
A projectile is launched from and returns to ground level. Air resistance is absent. The horizontal range of the projectile is R = 175 m, and the horizontal component of the launch velocity is Vox = 25 m/s. Find the vertical component Voy of the launch velocity.
|
Alright man, I looked up 2 dimensional kinematics and I'm still not sure how to solve this problem. My best guess:
Using the equation: y=v*t+.5*a*t^2
where:
y = vertical displacement
v= velocity
t= time
a = acceleration
and figuring out that t=7
you have:
0 = v*7+.5*(-9.8)*7^2
which got me
v= 34.3 m/s
have no idea if that's actually right though