I'm not really sure how to use the algebra to prove the monkey shouldn't have dropped. Obviously if the hunter aimed the gun directlt at the monkey it wouldn't hit it as gravity would pull it down.
A huntsman fires a gun pointing at a monkey hanging on the branch of a tree. The huntsman and the base of the tree are on a flat horizontal. At the moment the huntsman fires the gun, the monkey will release it's grip on the branch of the tree. Use the initial bullet velocity Vo, horizontal distance Dx, monkey height Dy, and inital trajectory angle above.
a)Show the monkey shouldn't have dropped
b)For what minimum bullet velocity? Vo = (Dx/cos0)(sqrt(g/2Dy))
So I need to show that the monkey shouldn't have dropped when the bullet was fired somehow.... And then I have to find what minimum bullet velocity.
Thank you so much guys..
I have to do this with only variables...
a. the distance the monkey drops in time t, is the same distance the bullet drops in time t. You don't have to know t to know that.
b. you have the formula there.
That velocity is the velocity it takes for the bullet to reach impact directly below the monkey.
But isn't the bullet going up ? I'm so confused on how to write this with variables... So Dy = T x-9.8?
If there was no gravity, the monkey would not fall, nor the bullet fall from its path. But gravity exists, and both fall from the tree, and the straight line path, at the same rate.
I know that ss the bullet passes through the air, it will arc down due to gravity. If the monkey held the branch, and the hunter was far enough away, the bullet might pass beneath him, but because the monkey is falling while the bullet is travelling, he's going to get shot.... But how do I show this with algebra?
And thank you so much for taking the time to help me.
easy.
the impact will be directly below the monkey orig position, figure how long it takes the bullet to go that horizonal distance (time=ditance/horizonalcompoentvelocity)
Now, using that time, show the y position of the bullet is exactly the original height of the monkey-how far it fell.
have fun, it's just algebra.
Bullets horizontal displacement is Dx = (VoCos0)((VoSin0/9.8m/s^2x)2)
I'm not sure how to fund the y position of the byllet is exactly the original height of the monkey - how fat it fell..?
The bullet must go horizontally a distance, Dx, so Dx = Vh / t or t = Vh / Dx = VcosA / D and
Vup = VsinA + gt .... or t = (Vup -VsinA) /g .... sub
VcosA /D = (Vup -VsinA) /g .... V = D(Vup -VsinA)(g cosA)
To show that the monkey shouldn't have dropped, we need to consider the motion and trajectory of both the bullet and the monkey. First, let's analyze the bullet's trajectory.
1. Bullet trajectory:
The bullet's velocity can be decomposed into horizontal (Vx) and vertical (Vy) components. The horizontal component remains constant throughout the bullet's flight, while the vertical component is affected by gravity. The bullet's trajectory can be described by the equation:
x = Vx * t (1)
y = Vy * t - (1/2) * g * t^2 (2)
where x and y represent the horizontal and vertical distances traveled by the bullet, respectively, t is the time, and g is the acceleration due to gravity.
2. Monkey's motion:
When the monkey releases its grip on the branch, it will also follow a projectile motion. The monkey's vertical motion is similar to that of the bullet, as it is also subject to gravity. However, the monkey's horizontal motion is different from the bullet's because the monkey does not have an initial horizontal velocity component.
Using the initial bullet velocity Vo, horizontal distance Dx, and the monkey height Dy, we can establish the relationship between the bullet's trajectory and the monkey's motion. The bullet will hit the monkey if its horizontal position (x) matches the horizontal distance to the monkey (Dx), and the bullet's vertical position (y) matches the monkey's height (Dy). Thus, we can write:
Dx = Vx * t (3)
Dy = Vy * t - (1/2) * g * t^2 (4)
3. Combining equations:
To solve for t, we can equate equations (1) and (3), and then substitute this value of t into equation (2) and (4):
Vx * t = Dx (5)
Vy * t - (1/2) * g * t^2 = Dy (6)
So we have two equations (5 and 6) with two unknowns (Vx and Vy). To find the minimum bullet velocity, we will need to find the values of Vx and Vy that satisfy these equations.
Note: The equation for minimum bullet velocity Vo, given in your problem, is a result of solving equations (5) and (6) and finding the minimum value needed to ensure the bullet hits the monkey.
Once we have determined the minimum value for Vo, we can conclude that the monkey shouldn't have dropped for any lower bullet velocity.