a steel ball rolls with a constant velocity across a tabletop 0.91 m high. it rolls off the table and hits the ground 0.77 m from the edge of the table. how fast was the ball rolling?
The time it takes the ball to fall a distance h = 0.91 m is
t = sqrt(2h/g) = 0.431 s
The horizontal velocity component while rolling (and while falling) remains
Vx = 0.77 m/0.431 s = 1.79 m/s
To find out how fast the steel ball was rolling, we can use the principles of projectile motion. We'll need to consider the vertical and horizontal components of the ball's motion separately.
First, let's find the time it takes for the ball to fall from the table to the ground. We can use the equation for vertical motion:
y = ut + (1/2)gt^2
Where:
y = vertical displacement (0.91 m)
u = initial vertical velocity (0, since the ball is rolling horizontally)
t = time
g = acceleration due to gravity (-9.8 m/s^2)
Since the ball rolls off the table horizontally, the time it takes to hit the ground is the same as the time it would take for an object in free fall to reach the same vertical displacement. We can ignore the initial vertical velocity (0) because it cancels out.
0.91 = (1/2)(-9.8)t^2
Rearranging the equation and solving for t:
t^2 = (2 * 0.91) / 9.8
t^2 ≈ 0.1863
t ≈ 0.4317 seconds
Next, let's find the horizontal component of the ball's motion. We know that the horizontal distance is 0.77 m and the time is 0.4317 seconds. We can use the equation:
x = ut + (1/2)at^2
Where:
x = horizontal displacement (0.77 m)
u = initial horizontal velocity (unknown)
t = time (0.4317 s)
a = horizontal acceleration (0, since there is no horizontal force acting on the ball)
Simplifying the equation:
0.77 = u * 0.4317
Solving for u:
u ≈ 1.786 m/s
Therefore, the ball was rolling at approximately 1.786 m/s.