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March 29, 2017

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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!! - ,

    Thank you so much for taking the time to help me

  • Physics question - Damon please help!! - ,

    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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