Interesting physics problem, please help?

Question

Tom the cat is chasing Jerry the mouse across the surface of a table 1.0 m above the floor. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of 4.8 m/s.

Where will Tom strike the floor?
(I calculated this and got 2.16 m from the table)

What velocity components will he have just before he hits? (Use a coordinate system in which up is positive.)
(I calculated this and got 4.77 m/s in the x direction)

___ y direction m/s (for some reason, I can't figure this one out)

GREG G!! how did u get that?! could someone please explain it to me, showing me step by step solutions?

anyone want to help with this one?

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 21.0 m/s. The cliff is h = 42.0 m above a flat horizontal beach.

How long after being released does the stone strike the beach below the cliff?
__ s
With what speed and angle of impact does the stone land?
___ m/s
____° below the horizontal

Additional Details

5 minutes ago
Oh, and I got 2.93 secs for the first question, if that helps any.

Answer 1

1. First of all let me say that this is a bad question and it really does not have a unique solution. This is so as the physics involved regarding motion of objects is valid only for inanimate objects which do not themselves exert any force. The cat is alive. It is going to struggle and exert force from within. It may even jump upwards. This complicates the problem. So, to solve, we assume, the cat will fall off like a dead cat or inanimate object.

2.(a). With the above assumption, initial velocity, just off the table surface is u = 4.8 m/s (assume in the horizontal, X direction)
Vertical height, distance, S = 1.0 m (given)
Acceleration = acceleration of gravity = 9.8 m/s2 (assumed).
To calculate the horizontal distance traveled, first calculate the time of fall, t.
Formula to be used, S = ut + 1/2 a t2

For vertical component of the motion, S= y = 1.0 m; u = Uy = 0.
So, y = 0 +4.9 t2; t = sq-rt y/4.9 = sq-rt 1/4.9 = 0.452 seconds.

Now consider the horizontal component of motion.
Ux = u = 4.8 m/s, a = ax =0; S = Sx =?
Applying the same formula, S = X = ut = 4.8x0.452 = 2.1696 m.
This may be approximated to 3 significant figures.
Then, X = 2.17 m.

The cat will land on the floor, 2.17 m from the edge of the table(measured along the floor).

2(b). Velocity component. Since we assume, the acceleration is only due to gravity, it is along the vertical direction and its horizontantal component is zero. So, ay = -9.8 m/s2 (negative for downward, as we assume up is positive).
So, the horizontal component of the motion is uniform; the velocity remains constant. vx = ux = u =4.8 m/s.

The vertical component: Initial velocity, uy = o;
Using the formula v = u +at;
vy = 0 + -9.8x0.452 = - 4.4296 m/s.
Again, approximating to 3 sf, vy = -4.43 m/s.
The negative sign shows, it is downward.
(we assumed, up is +ve).

Answer 2

First of all, his component in the horizontal (x) direction doesn't change from 4.8 m/s

The only thing you need to know is how fast was he going in the y direction.

He fell 1 meter. 1 = 1/2at^2
t = sqrt(1/4.9) is the time it took him to fall.

v = at = 9.8 * sqrt(1/4.9) is his velocity falling or in the y direction.

You can also get how far from the table he hit by using the same time it took him to fall and multiplying by the horizontal velocity.

I get sqrt(1/4.9) * 4.8 m/s as being the distance from the edge of the table.

Since the problem tells you to use 'Up' as the positive y direction, your velocity in the y direction will be negative.

Answer 3

If Tom the cat is going 4.8 m/m second when he flys of the table than the speed should be the same since he is falling (y direction.)

Answer 4

I'm not very good at physics so I apologize if this is a bit silly, but since he is falling off the table the horizontal (x), velocity would be due to his running on the table, would the vertical (y) not be due to the force of gravity?

Answer 5

I don't know. Watch the cartoon and see.

Answer 6

4.43 m/s downward