What is Jane and Cheetah's Speed Before Grabbing Tarzan?

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In summary, Tarzan's friends Jane and Cheetah have to decide whether to bail him out of trouble once again. To do so, they need to swing down on a vine and grab him from a cliff 10m below and then reach a perch 5m above him. Using the conservation of energy principle, it can be determined that their total energy must be greater than 6867 Joules in order to safely reach the perch. Since their combined mass is 70kg, their speed just before grabbing Tarzan must be calculated by solving the equation 35V2= 6867.
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Tarzan is in trouble once again and Jane and Cheetah have to decide whether to try to bail him out. Jane and Cheetah (70kg combined) are perched on a cliff 10m above Tarzan (70kg) and can swing down on a vine and grab him. To get to safety, they need to get to a perch 5m above Tarzan on the other side.

Determine the speed of Jane and Cheetah just before they would grab Tarzan. Assume no air drag or friction.
 
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  • #2
That should be fairly straight forward if you use "conservation of energy". Take Tarzan's position as 0 potential energy. Since they need to get to a position 5 m above Tarzan, where the potential energy of all 3 is 140g(5)= 700(9.81)= 6867 Joules, their total energy must be greater than 6867 Joules.
You know that the combined mass of Cheetah and Jane is 70 kg. so, taking V as their speed (in m/s) just before they "pickup" Tarzan, their kinetic energy is 35V2 Joules. Since their potential energy is now 0, their total energy is 35V2.

Solve 35V2= 6867.
 
  • #3


To determine the speed of Jane and Cheetah just before they grab Tarzan, we can use the equation for conservation of energy. Since there is no air drag or friction, all of the initial potential energy from being perched on the cliff will be converted into kinetic energy as they swing down on the vine.

The initial potential energy of Jane and Cheetah is given by mgh, where m is their combined mass of 70kg, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the cliff (10m).

PE = mgh
= (70kg)(9.8 m/s^2)(10m)
= 6860 Joules

This potential energy will be converted into kinetic energy, given by the equation KE = 1/2mv^2, where v is the velocity.

KE = 1/2mv^2
6860 = 1/2(70kg)v^2
v^2 = (2)(6860)/70
v = √196 m/s
v = 14 m/s

Therefore, the speed of Jane and Cheetah just before they grab Tarzan will be approximately 14 m/s. This is assuming they have a perfectly efficient swing and all of their potential energy is converted into kinetic energy. If there are any inefficiencies or losses, the actual speed may be slightly lower. But with this speed, they should be able to reach the perch 5m above Tarzan on the other side and save him from trouble.
 

FAQ: What is Jane and Cheetah's Speed Before Grabbing Tarzan?

What is "Tarzan in trouble - Kinematics"?

"Tarzan in trouble - Kinematics" is a physics problem that involves analyzing the motion of Tarzan as he swings on a vine across a river to rescue Jane.

What is kinematics?

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion.

How do you solve the problem of "Tarzan in trouble - Kinematics"?

To solve this problem, you would need to use equations of motion, such as the equation for vertical motion and the equation for horizontal motion. You would also need to consider the initial velocity, acceleration due to gravity, and the distance between the two points where Tarzan starts and finishes his swing.

What information do you need to solve "Tarzan in trouble - Kinematics"?

To solve this problem, you would need to know the initial velocity of Tarzan, the height of the tree he is swinging from, the height of the cliff that Jane is hanging from, and the distance between the two points.

Why is "Tarzan in trouble - Kinematics" an important problem in physics?

This problem demonstrates the principles of motion and helps us understand the mechanics of swinging and projectile motion. It also helps us develop problem-solving skills and apply equations of motion in a real-life scenario.

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