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April 20, 2015

April 20, 2015

Posted by **physics** on Thursday, December 25, 2008 at 3:13pm.

I'm having some problems here with this

for simpicity issues

i said launch at .2m off the ground at origin and reach a height of 1.1m at 1m away just incase something screws up.

I need to find the

Vi

Vyi

Vxi

and am having some trouble and dont know what angle is best so that less speed is needed for this to work

- physics -
**drwls**, Thursday, December 25, 2008 at 4:14pmYou have selected 0.2 m for the initial height and 1.1 m for the maximum height. That allows you to calculate the time the projectile must spend coming down (t2) and going up (t1).

(g/2) t2^2 = 1.1 m

t2 = 0.474 s

(g/2) t1^2 = 0.9 m

t1 = 0.429 s

Initial y component of velocity:

Vyi = g*t1 = 4.204 m/s

Total time of flight = 0.903 s

Vxi*0.903 s = 2.0 m

Vxi = 2.214 m/s

Vi = sqrt(Vxi^2 + Vyi^2) = 4.751 m/s

These two components determine the initial launch angle as

arctan(Vyi/Vxi) = 62.2 degrees

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