Here's a question for you brainiacs out there!?

Question

Tarzan, who has a mass of 60.0 kg, runs toward a lake at a speed of 5.00 m/s. He grabs a massless rope of length 10.0 m and begins to swing out over the lake, which is 20.0 m below. Taking t = 0 to be when he grabs the rope, at what time should he release the rope so he travels the greatest distance before hitting the lake? Assume the rope hangs just even with the edge of the cliff.

Bonus Question: What angle with the vertical does the rope make at this time?

I made this question up, and I'm pretty sure it has all of the needed information to solve it.

Answer 1

First, the mass is irrelevant.

Second, did you just sit down and think, "hmm, what's the hardest 4A type problem I could come up with"?

Conservation of energy tells us both the maximum angle he would go up (about 29°), as well as give us a relationship between the speed and the angle.

Key to using conservation of energy is by looking at the geometry. If h is the height term in the potential energy and y is the distance from position up to the level of the knot holding the rope, then
h+y=l
where l=length of the rope.

This gives us the expression
h=l(1-Cos[θ])

From there we can get an expression for the velocity in both the x and y directions.

From the velocity in the y direction and the height (20 m+the height from swinging upwards) we can get the time to hit the lake. Use the kinematic equation

y=
y_i+(v_y i)t+½at²

and the quadratic formula to find the time.

From the velocity in the x direction and the time we can get the distance traveled after swinging out. Don't forget to add the distance swung out to the previous x distance.

Now you have x as a function of θ.

Perhaps someone more tenacious than myself could then take the derivative of that function, set it to zero and solve for the angle by hand.

But, for a problem like this, Mathematica is your friend.

Use the FindMaximum function to find the maximum value of x as a function of θ.

Using the initial speed and the final speed (if you know the angle, then you know the final speed) you can estimate the average speed. Since the initial speed and final speed are close, the average should be close to

v{average}
=(v{initial}+v{final})/2

The distance is just the angle in radians times the radius.
s=rθ

Time, then, is distance/speed.
t=s/v

You could get an exact answer for the time, by taking the integral

t=∫(1/ω)dθ
ω=v/r=v/l

but that sounds like more work, and the answer you'll get will be close to the approximation I mentioned.

Answer 2

When he grabs the rope, he will have a kinetic energy of
Ek = mv²/2 = 60*5²/2 = 750J
As he swings up on the rope (we must assume the rope is hanging vertically, since it isn't specified) he will reach a height at which his velocity is zero (when all of his kinetic energy has become potential energy). This will be at
mgh = 750J
h = 750/(mg) = 750/(60*9.8) = 1.275 meters. At this point, he is 20+1.275 = 21.275 meters above the lake, and
10sin(arccos(10-1.275)/10) = 6.203 meters horizontally towards the lake from the top of the cliff.

HTH ☺

Doug