How do you calculate speed without taking mass into account?

[GregoryG.]

Homework Statement

During the World Series a ball is hit straight up from the batter’s box and reaches a height of 295 ft. Neglecting air friction, what speed (mph) will the baseball attain if it is caught by the catcher at the same height it was hit?

Homework Equations

The equation I pull out of this is the one for gravitational potential energy

Ug=M*G*H
G = 9.81 m/s
H = 295 feet = 89.9 m

It could also have something to do with the ball having no kinetic energy when it make contact with the bat, then having no gravitational potential energy when the catcher catches it but, I can't figure out how to put these two formulas together.

Uk=(1/2)*mv^2
3. An attempt at the problem


Ug = 9.81 m/s * 89.9m
Ug = 882m/s = 2030m/h (seems unlikely this ball will break mach 1)

This would only work if we disregard mass. We've done this in other problems, but it seems very important here.

 

[Mister T]

What is the kinetic energy of the ball when it's at the highest point in its trajectory?

 

[SteamKing]

Homework Statement

During the World Series a ball is hit straight up from the batter’s box and reaches a height of 295 ft. Neglecting air friction, what speed (mph) will the baseball attain if it is caught by the catcher at the same height it was hit?

Homework Equations

The equation I pull out of this is the one for gravitational potential energy

Ug=M*G*H
G = 9.81 m/s
H = 295 feet = 89.9 m

Why did you choose this particular equation? It doesn't help you to answer the primary question, How fast will the ball be traveling when it is caught?

It could also have something to do with the ball having no kinetic energy when it make contact with the bat, then having no gravitational potential energy when the catcher catches it but, I can't figure out how to put these two formulas together.

Uk=(1/2)*mv^2

You are making this problem much harder than it truly is.

If you drop something from the top of a building which is 295 feet high, how fast will that object be traveling when it hits the ground?

What if I told you the name of the player who hit the ball was Manny Suvat?

3. An attempt at the problem


Ug = 9.81 m/s * 89.9m
Ug = 882m/s = 2030m/h (seems unlikely this ball will break mach 1)

What are the units of g * h? (Hint: it ain't m/s).

This would only work if we disregard mass. We've done this in other problems, but it seems very important here.

Since you aren't given the mass of the baseball, that should be a big old hairy hint that PE or KE will not be very useful to answering the question here.

 

[krebs]

Have you taken calculus?

 

[krebs]

We know that $V = V_0 - 9.8 t$ What is the antiderivative of this formula? Do it on paper and keep track of your units.

 

[gneill]

Homework Statement

During the World Series a ball is hit straight up from the batter’s box and reaches a height of 295 ft. Neglecting air friction, what speed (mph) will the baseball attain if it is caught by the catcher at the same height it was hit?

Homework Equations

The equation I pull out of this is the one for gravitational potential energy

Ug=M*G*H
G = 9.81 m/s
H = 295 feet = 89.9 m

Check your units for G. It's an acceleration, not a velocity. Also, the conventional variable for gravitational acceleration is "g" (lower case). Capital G is usually reserved for Newton's gravitational constant.

It could also have something to do with the ball having no kinetic energy when it make contact with the bat, then having no gravitational potential energy when the catcher catches it but, I can't figure out how to put these two formulas together.

Uk=(1/2)*mv^2
3. An attempt at the problem


Ug = 9.81 m/s * 89.9m
Ug = 882m/s = 2030m/h (seems unlikely this ball will break mach 1)

This would only work if we disregard mass. We've done this in other problems, but it seems very important here.

If you fix your units for "G" (should be "g"), then you'll see that the results of Ug would be $m^2/s^2$, which is equivalent to $J/kg$ (that's Joules per kg). This is not a bad thing! (see below)

Conservation of energy will work here! If you don't know the mass you can leave it as a symbol and it will cancel out by the end if it's not an essential quantity for the problem.

In fact, it is sometimes practical to work with what are called "specific" quantities for energy related problems. This is particularly true in fields like astrodynamics where the masses of objects like comets or asteroids are not well known. A specific energy is the energy per unit mass, so the units would be J/kg for example. So we could write:

$PE_s = g h$
$KE_s = \frac{1}{2} v^2$

 

[GregoryG.]

What is the kinetic energy of the ball when it's at the highest point in its trajectory?

It would be zero right? The ball isn't moving anymore because of the energy imparted to it by the bat at the highest point of its trajectory.

 

[GregoryG.]

Have you taken calculus?

I'm in Calc one now. We haven't discussed anti-derivatives, and this problem is from an intro algebra-based physics class. I don't think my teacher wants us to use Calculus.

 

[krebs]

I'm in Calc one now. We haven't discussed anti-derivatives, and this problem is from an intro algebra-based physics class. I don't think my teacher wants us to use Calculus.

Oh, sorry. Use the energy formulas then:
$$PE = gh$$
$$KE = \frac{1}{2}v^2$$

There is one point where the ball is not moving. That's when you can use the potential energy formula...

 

[Mister T]

It would be zero right? The ball isn't moving anymore because of the energy imparted to it by the bat at the highest point of its trajectory.

Right, so the total energy when the ball is that point (remember, total energy equals kinetic energy plus potential energy) would be $mgh$.

When the ball is at its lowest position the potential energy is zero, so the total energy is ... ?

Set the two amounts of energy equal to each other. (When you do that $m$ cancels.)

 

[krebs]

Right, so the total energy when the ball is that point (remember, total energy equals kinetic energy plus potential energy) would be $mgh$.

When the ball is at its lowest position the potential energy is zero, so the total energy is ... ?

Set the two amounts of energy equal to each other. (When you do that $m$ cancels.)

Not to nitpick technicality, but Gregory, do keep in the back of your mind that you are calculating velocity, KE and PE relative to the ground.