The simulation shows the classic physics situation of a monkey and hunter. In th

Question

The simulation shows the classic physics situation of a monkey and hunter. In this case, the hunter is trying to tranquilize the monkey, so the monkey can be relocated to a better habitat. The monkey is clever-when the tranquilizer dart (in blue) leaves the gun, the monkey (in purple) lets go of the tree branch, and starts to fall straight down. Note that there is a net (not shown) at the bottom to catch the monkey, so the monkey won't get hurt. The hunter anticipates the monkey's clever move and aims below the monkey to try to hit it. Being experts in physics now, we know this is not the right thing to do. Thus, the dart ends up passing some distance below the monkey. Although you can't use the exact values below with the simulation, the simulation should help you figure out whether your methods make sense. Let's say that the dart is launched with initial velocity components of 38.0 m/s horizontally and 9.880 m/s vertically. Imagine that this is taking place on a planet where the acceleration due to gravity is g 9.00 m/s, and neglect air resistance. (a) What is the distance between the dart and the monkey at the instant the dart passes directly below the monkey?

Explanation / Answer

given, initial; velocity components of the muzzle velocity of the bullet

Vx = 30 m/s

Vy = 9.88 m/s

acceleration due to gravity = g = 9 m/s/s

let say the dart takes time t to pass the monkey

then horizontal distance travelled by dart in time t x = Vx*t

for vertical distance, y = Vy*t - 0.5gt^2

y = Vy*x/Vx - 0.5g(x/Vx)^2

also, if monkey is at height h

vertical distance travelled by monkey y' = 0.5gt^2 = 0.5*g*(x/V x)^2

vertical distance between the dart and the monkey = h - y - y'

h - y - y' = h - Vy*x/Vx + 0.5g(x/Vx)^2 - 0.5*g*(x/V x)^2

h - y - y' = h - Vy*x/Vx = h - 0.329x

where x is horizontal distance form the monkey and h is initial height of the monkey

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