- 24 Oct '07 02:04

Yes, ignoring air resistance. That's what is messing me up, gravity. I don't even know what a high jumper is, with a constant acceleration of -5 m/s/s.*Originally posted by Ramned***To clarify, are we ignoring air resistance and gravity? Is he falling at a constant velocity of 5 m/s**

I think I will just skip that problem... - 24 Oct '07 02:06 / 1 editThis one is straightforward

**As a baseball is being caught, its velocity goes from 30.5 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.145 kg.**

(a) What is the baseball's acceleration?

So Acceleration = Sum of Forces/Mass

But how do you calculate the sum of the forces. I was thinking of using a motion equation but I can't get it to work. - 24 Oct '07 03:13Well the wording seems wrong, because you don't have information on the acceleration, only the decel, which means it should have said what is the deceleration. Anyway, 30.5 m/s in 5 milliseconds, 1/200th of a second seems to me just 200*30.5 or about 6100m/s/s. In G's, about 600 G's. Don't see how you could get that much decel in the real world with a baseball. I wonder how much accel there is when the ball is hit by a bat? Just a finger in the air judgement, but it seems like 30 M/S sounds like a reasonable velocity for a well hit baseball but unless you know how much time the bat and ball are in contact, you don't know the acceleration.
- 24 Oct '07 08:59

Deceleration is just negative acceleration. There's nothing wrong with the wording. That's why the answer should be negative.*Originally posted by sonhouse***Well the wording seems wrong, because you don't have information on the acceleration, only the decel, which means it should have said what is the deceleration.** - 24 Oct '07 09:05 / 1 edit

The kinetic energy of the jumper when he hits the pit is mv^2/2. This is reduced to zero by the force.*Originally posted by Drew L***A high jumper, falling at -5.0 m/s, lands on a foam pit and comes to rest, compressing the pit a distance of 0.59 m. If the jumper experiences an average net force of +1200 N while having his fall broken by the pit, what is the jumper's mass?**

___kg

The work done by the force is F x d, if F is the average total force.

So m = 2Fd/v^2

= 56.64 kg (57kg to 2s.f.) - 24 Oct '07 13:26 / 1 edit

isnt this a matter of V / T - to find acceleration upon contact? So you do not need mass, unless it wants to find force later.*Originally posted by Drew L***This one is straightforward**

[b]As a baseball is being caught, its velocity goes from 30.5 m/s to 0.0 m/s in about 0.0050 s. The mass of the baseball is 0.145 kg.

(a) What is the baseball's acceleration?

So Acceleration = Sum of Forces/Mass

But how do you calculate the sum of the forces. I was thinking of using a motion equation but I can't get it to work.[/b]

-30.5 / .0050 = a

then, f = .145(a)

a = -6100.

F = ma

F = .145(-6100)

F = -884.5 N

Pretty easy. Is this intro physics class or something?

Meanwhile, for those looking for some hard-time physics try to solve mine in Master of Physics thread

To make this one bit harder (but pretty basic still)

A baseball is pitched at 101.0 miles/ hour, thrown horizontally. How far would the ball fall vertically by the time it reached home plate, 60.5 feet away. (*Hint: Convert to meters)? - 28 Oct '07 01:00

Wouldn't that depend on whether it was pitched by Cy Young V Shaq O'neal?*Originally posted by Ramned***isnt this a matter of V / T - to find acceleration upon contact? So you do not need mass, unless it wants to find force later.**

-30.5 / .0050 = a

then, f = .145(a)

a = -6100.

F = ma

F = .145(-6100)

F = -884.5 N

Pretty easy. Is this intro physics class or something?

Meanwhile, for those looking for some hard-time physics try to solve mine ...[text shortened]... l fall vertically by the time it reached home plate, 60.5 feet away. (*Hint: Convert to meters)?

6 feet up V 7 feet up seems to me would make a difference and no vertical height is given. What if it is pitched by a midget? So 4 feet up V 7 feet.

Also you can only start out horizontal, the ball would immediately start a parabolic curve downwards, so at best, if the ball was thrown in outer space with no force downwards, it would take 0.409 seconds to reach home plate.

So the answer depends on the initial pitching height. S=(A(T^2))/2.

Keeping it in feet, 60 MPH is 88 Feet/sec so 101 MPH is (101/60)*88

so that is 1.6833*88 or (148.13 F/S)/60.5 feet = 0.409 seconds and

so S=(32*.167)/2, it would drop 2 2/3 feet in that time, doesn't matter if it is dropped straight away or pitched, it still drops at 32 F/S^2 but it sure matters what height it was pitched from if it is to land somewhere near the catchers mitt. Suppose it was pitched by an 8 foot dude and he let go at the top of his swing and it was therefore 12 feet in the air when pitched, it would be just under 10 feet in the air as it whizzed by home plate, making it near impossible to catch. If it was pitched by a 6 footer and he let loose at shoulder height then it would be about 5 feet up and would end up at about half that, a perfect height to be caught by the catcher. Why is it a hint to do the arithmetic with metric?