# How high will he swing off the ground?"

• Moogle

#### Moogle

Here is a question I had to figure out in my grade 11 physics class.

"If a 40 kg child is running at a rope suspended from the branch of a tree at 8.0 m/s, and then jumps and grabs it, how high will he swing off the ground?"

(I had to translate it so pardon me if it is hard to understand)

One of my friends suggested doing Height=(M)/(V)/(A) but he has a habit of messing up.
I would appreciate any help from anyone.
Thx.

First of all, welcome to the forums !
Your question is clear enough
Normally in the HomeWork help forum, we provide help to a certain problem being faced by the members, and do not provide ready-answers.
But ... anyway ..

The best way i see this question can be solved is through conservation of energy.
Conservation of energy means that the whole energy of a system at a certain time, will be equal to the whole energy of it at another time.
We can apply (the simple form of) the law of conservation of energy when the system is conservative (In most cases this means that there is no air resistance, no friction ... etc ..)
In your example, we will omit the air resistance and friction to get an easy answer.

The energy of the system before the child grabbed the rope (E1) is its Kinetic energy (KE1) added to its Potential energy (PE1).
E1=KE1+PE1
Same for the energy at the heighest point after he grabs the rope
E2=KE2+PE2

Now, let's see a little more about KE and PE.
KE=0.5*m*v2
Where (m) is the mass, and (v) is the velocity (if you have no idea what the velocity is, think of it like 'speed' for the moment, although it is not exactly speed).
PE=m*g*S
Where (m) is the mass, (g) is the acceleration due to the Earth's gravity, and (S) is the displacement between the object and a reference point (the Earth's surface normally).
So :
E1=0.5*m*v12+m*g*S1
But since S1 is 0 (the boy was at the surface of Earth at the begining), then :
E1=0.5*m*v12

Now let's see E2
E2=0.5*m*v22+m*g*S2
But since v2 equals 0 at the topmost point, then :
E2=m*g*S2

Remmember that from the coservation of energy:
E1=E2

Can you take it from here and continue alone ?

I hope i helped.

If the kid is running @ 8 m/s it must be the fastest kid in the world.

Thats like about as fast as a bullet/rocket.

Reminds me when a idiot tutor in college says: "Now image that's this conveyor belt (which had a box on it) is traveling at say ... 50 m/s.", I burst out laughing, and he said "yeah its going like a bullet let's change that 25 m/s". At which I laughed alloty harder.

Originally posted by Dave_3of5
If the kid is running @ 8 m/s it must be the fastest kid in the world.

Thats like about as fast as a bullet/rocket.

Reminds me when a idiot tutor in college says: "Now image that's this conveyor belt (which had a box on it) is traveling at say ... 50 m/s.", I burst out laughing, and he said "yeah its going like a bullet let's change that 25 m/s". At which I laughed alloty harder.

Once I was a TA for a class in which the professor (!) had a problem in km/h which turned out to be something like 8 times the speed of light.

Um...yes 8 m/s might be a BIT fast for a kid...but a bullet goes 100 times faster...

Are we talking about meters per second here?

Huh? 8 m/s is 28.8 km/h. I promise you, a bullet travels much faster than that, not to mention light.

8m/s is approximately 18 miles per hour. human adults can run over 25 miles per hour. While this number is fast for a child, it is probably possible in a short distance, all-out sprint. Although 5 or 6 m/s would be more reasonable, for the sake of the physics problem it is within reason.