Thread: ball rolls off a table

Question

A ball rolls off a table 4ft high while moving at a constant speed of 5ft/s.

a) How long does it take for the ball to hit the floor after it leaves the table?

b) At what speed does the ball hit the floor?

c) If a ball were dropped from rest at table height just as the rolling ball leaves the table, which ball would hit the ground first? Justify your answer.

How should I solve this question? Thank you very much!

Originally Posted by Jenny20

Question

A ball rolls off a table 4ft high while moving at a constant speed of 5ft/s.

a) How long does it take for the ball to hit the floor after it leaves the table?

b) At what speed does the ball hit the floor?

c) If a ball were dropped from rest at table height just as the rolling ball leaves the table, which ball would hit the ground first? Justify your answer.

How should I solve this question? Thank you very much!

a) How long does it take for the ball to hit the floor after it leaves the table?

Imagine the rolling ball as a projectile fired horizontally at 5 ft/sec from a height of 4 ft from the floor.

For the vertical component of the velocity of the ball:
H = (Vo_y)*t -(1/2)gt^2
where
H = height of the ball at any time t, measured upwards, in ft.
Vo_y , read Vo sub y, = initial vertical velocity, going upwards, in ft/sec
g = acceleration due to gravity is, say, 32 ft/sec/sec
t = time for the ball to rise to H, in seconds.

But there is no vertical component of the initial velocity. The initial velocity is all horizontal,
S0,
-4 = 0*t -(1/2)(32)t^2 --------the H is "negative" because we're measuring it dowhwards.
t^2 = 4/16 = 14
t = 1/2 sec. ------------time for the ball to hit the floor, answer.

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b) At what speed does the ball hit the floor?

V = Vo_y -gt
V = 0 -32(1/2) = -16 ft/sec ----------answer.
Negative, because it is going downwards.

For the horizontal component of the ball's initial velocity:
D = (Vo_x)*t
where
D = horizontal distance travelled by the ball after time t, in ft.
Vo_x = horizontal component of the initial velocity, in ft/sec.
So when t = 1/2 sec,
D = 5*(1/2) = 2.5 ft.

----------------------------------
c) If a ball were dropped from rest at table height just as the rolling ball leaves the table, which ball would hit the ground first? Justify your answer.

None. Both balls will hit the floor at the same time. Because,
The velocity of the ball when it hits the floor is given by
V = Vo +at = Vo +gt = 0 +32t = 32t ft/sec -----here we assume going downward is positive.
H = (1/2)(Vo +V)t
4 = (1/2)(0 +32t)t
4 = 16t^2
t^2 = 4/16 = 1/4
t = 1/2 seconds ----same as when the ball was rolling at 5ft/sec.

I need to add two comments to ticbol's answer to part b.

The first is a technical comment:

V = Vo_y -gt

This is the equation for the vertical component of the velocity. With a coordinate system defined such that vertical upward is the positive direction, this answer will indeed be negative. But the question is asking for the speed, which is the magnitude, or size, of the velocity. This number can never be negative.

The second thing is that the ball has a horizontal component to its velocity as it hits the floor, since it had one when it left the table. (There is nothing in the horizontal direction to impede the motion, so there will be no acceleration in the horizontal direction.)

So we have a horizontal component of velocity of vh = 5 ft/s and a vertical component of vv = -16 ft/s. To find the speed (magnitude of the velocity) we need to add these two vectorally. You can draw the picture and, as usual, the components form a right triangle with the velocity vector. So the speed will be:
v = sqrt{vh^2 + vv^2} = sqrt{(5)^2 + (-16)^2} = 16.763 ft/s

-Dan

Originally Posted by topsquark

I need to add two comments to ticbol's answer to part b.

The first is a technical comment:

This is the equation for the vertical component of the velocity. With a coordinate system defined such that vertical upward is the positive direction, this answer will indeed be negative. But the question is asking for the speed, which is the magnitude, or size, of the velocity. This number can never be negative.

The second thing is that the ball has a horizontal component to its velocity as it hits the floor, since it had one when it left the table. (There is nothing in the horizontal direction to impede the motion, so there will be no acceleration in the horizontal direction.)

So we have a horizontal component of velocity of vh = 5 ft/s and a vertical component of vv = -16 ft/s. To find the speed (magnitude of the velocity) we need to add these two vectorally. You can draw the picture and, as usual, the components form a right triangle with the velocity vector. So the speed will be:
v = sqrt{vh^2 + vv^2} = sqrt{(5)^2 + (-16)^2} = 16.763 ft/s

-Dan

Hey, you are correct on the second thing. I completely forgot about the gravity and the horizontal velocity acting together on the ball.

Boy, what a day it was yesterday!

Originally Posted by ticbol

Hey, you are correct on the second thing. I completely forgot about the gravity and the horizontal velocity acting together on the ball.

Boy, what a day it was yesterday!

Been there, doing that today.

-Dan