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..
Thank you so much for taking the time to help me
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)
To find the minimum speed for the bullet, you need to set up an equation using the given variables and then manipulate it to obtain the desired form.
Let's start with the equation you derived:
Dx = (Vo * cosθ) * (2 * ((Vo * sinθ) / g))
Now, let's rewrite this equation to solve for Vo:
Divide both sides of the equation by cosθ:
Dx / cosθ = 2 * ((Vo * sinθ) / g)
Multiply both sides of the equation by g:
g * (Dx / cosθ) = 2 * Vo * sinθ
Divide both sides of the equation by 2 * sinθ:
(g * Dx) / (2 * sinθ * cosθ) = Vo
Now, let's simplify the denominator:
Using the identity sin(2θ) = 2 * sinθ * cosθ, we can rewrite the denominator:
(g * Dx) / (2 * sinθ * cosθ) = Vo * (1 / sin(2θ))
Finally, let's substitute sin(2θ) = 2 * sinθ * cosθ:
(g * Dx) / (2 * sinθ * cosθ) = Vo * (1 / (2 * sinθ * cosθ))
(g * Dx) / (2 * sinθ * cosθ) = Vo * (1 / (2 * sinθ * cosθ))
Cancel out the common terms of sinθ and cosθ:
(g * Dx) / (2 * sinθ * cosθ) = Vo * (1 / (2 * sinθ * cosθ))
Vo = (g * Dx) / (2 * sinθ * cosθ)
Now, let's simplify the expression further:
Using the identity cosθ / sinθ = cotθ, we can rewrite the expression:
Vo = (g * Dx) / (2 * sinθ * cosθ)
Vo = (g * Dx) / (2 * sinθ * cosθ)
Vo = (g * Dx) / (2 * (sinθ * cosθ) / sinθ)
Vo = (g * Dx) / (2 * (sinθ * cosθ) / sinθ)
Vo = (g * Dx) / (2 * (sinθ * cosθ) / sinθ)
Vo = (g * Dx) / (2 * cotθ)
Vo = (g * Dx) / (2 * cotθ)
Since cotθ is equal to cosθ/sinθ, we can rewrite the expression again:
Vo = (g * Dx) / (2 * cotθ)
Vo = (g * Dx) / (2 * (cosθ/sinθ))
Vo = (g * Dx) * (sinθ / (2 * cosθ))
Now, let's simplify the expression even further:
Vo = (g * Dx) * (sinθ / (2 * cosθ))
Vo = (g * Dx) * (sinθ / (2 * cosθ))
Vo = Dx * (g / (2 * cosθ)) * sinθ
Finally, let's simplify the expression to the desired form:
Vo = (Dx / cosθ) * sqrt(g / (2 * (1 / sinθ)))
Vo = (Dx / cosθ) * sqrt(g / (2 * sinθ))
Vo = (Dx / cosθ) * sqrt(g / (2Dy))
So, the minimum speed of the bullet, Vo, can be derived in the desired form as Vo = (Dx / cosθ) * sqrt(g / (2Dy)).