Posted by Jennifer on Monday, November 11, 2013 at 12:44am.
I am doing the monkey and hunter question, and I need to find out what the minimum speed the bullet has to be..
The variables I have are g, Vo, theta, Dx(horizontal distance from monkey), and Dy (vertical distance from monkey).
I need to derive Vo in the form of (Dx/cos0)(sqrt(g/2Dy))...
I have managed to derive Vo = Vo = sqrt((0.5(Dx/cos0/sinO))g) from Dx = (VoCosθ)(2((Vo)(Sinθ))/g), but its not the one I need..
- Physics question - Damon please help!! - Jennifer, Monday, November 11, 2013 at 12:45am
Thank you so much for taking the time to help me
- Physics question - Damon please help!! - Damon, Monday, November 11, 2013 at 12:16pm
If the shot is fired as the monkey falls, fire exactly at the monkey because the bullet and the monkey will both accelerate down at g and meet.
Dy/Dx = tan T
Dy = Dx tan T
It does not matter what Vo is as long as the bullet and monkey do not hit the ground before colliding.
So for monkey:
Dy = (1/2) g t^2
so t = sqrt(2 Dy/g)
then for bullet
h = 0 when
0 = (Vo sin Theta) t - 1/2 g t^2
same t for both
(g/2) (2 Dy/g) = Vo sin Theta sqrt (2 Dy/g)
(g/2) sqrt(2 Dy/g) = Vo sin Theta
sqrt (g Dy/2) = Vo sin Theta
Vo = [ sqrt g Dy/2 ] / sin Theta
the very smallest Vo will be as theta approaches 90 degrees (shoot straight up from under tree)
In that case Vo = sqrt (g Dy/2)
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