Random Rants #XXIV - The Angry Mob

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Actually, if the projectile was ********, you'd also need to factor in the retardation rate.
 
I'm too lazy to actually do the calculations, but I think you'd go about it by knowing that the velocity at half max height is given by using Pythag's Theorem on the two component vectors at half max height. Use a variable to represent the launch angle. You could use -9.8t+v0 and it's integral, -4.9t^2+v0t, to get the time at which half max height was reached. You know that the horizontal component of the velocity will stay constant, so find the vertical component of the velocity at half max height and use Pythag's to find the velocity. Set the velocity equal to 3/4 of your initial launch velocity, and solve for your launch angle (probably using sin^2+cos^2=1 along the way).
 
Aren't the unknown quantities supposed to cancel out anyway?
 
There haven't been for a while actually. :)

Thank you wet season for not giving Chavez an excuse to cut power. :)

(Although rural areas still get them, sometimes for the whole day)
 
hey, I bet wolfram alpha can help you with physics
 
It is 3:09am, I have yet to fall asleep. I've pushed back my alarm as far as I dare do so, but it's still gonna go off in less than 4 hours. Tomorrow is gonna suck.
 
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