A monkey is hanging from a tree limb 10m above the ground. A wildlife biologist

Question

A monkey is hanging from a tree limb 10m above the ground. A wildlife biologist who is 20m away from the monkey horizontally throws a banana in the general direction of the monkey. The initial speed of the banana is 17.5m/s, and the velocity vector of the banana initially makes a 37 degree angle with the horizontal. At a time tm after the banana has been thrown, the monkey lets go of the limb, dropping straight down. (a) Write equations giving the x- and y- coordinates of the banana as functions of time, taking t = 0 to be the time at which the banana is released. (b) Write an equation giving the y- coordinate of monkey as a function of time, taking t = 0 to be the time at which the banana is released. (e) Find the time tm at which the monkey should release the branch in order to catch the banana in midair.

Explanation / Answer

Part A)

In the x direction

dx = vxt

dx = vcos(angle)t

If we want to put numbers in this, that is

dx = (17.5)(cos37)t = 14.0t

In the y direction

dy = vyt + .5gt2

dy = vsin(angle)t + .5gt2

If we want to put numbers here, we get

dy = 17.5(sin 37)t + .5(-9.8)(t2)

dy = 10.5t - 4.9t2

Part B)

For the monkey

dy = vyt + .5gt2

The monkey starts from rest

dy = .5gt2

dy = -4.9t2

Since the monkey drops at a time tm after the banana is thrown, the t for the monkey is t - tm

dy = -4.9(t - tm)2

Part C)

Now in the x direction for the banana.

dx = 14.0t

20 = 14t

t = 1.43 sec

Then for the banana in the y

dy = 10.5t - 4.9t2

dy = 10.5(1.43) - 4.9(1.43)2

dy = 4.99 m

For the monkey to fall 4.99 m

dy = -4.9t2

-4.99 = -4.9t2

t = 1.01 sec

Then 1.43 - 1.01 = .422 sec

The monkey needs to release .422 sec after the banana is thrown.

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