d = distance = 0.76 m
a = acceleration due to gravity = 9.81 m/s^2
u = initial velocity = 0 (as the ball rolls off the table the
vertical velocity = 0
t = time = missing so we need to solve it
So we use the equation d = ut + 1/2 at², and ever since u is
zero, ut is zero and the equation becomes to d = 1/2 at² and this reorders to t
= sqrt (2d/a) = 0.39 seconds.
Since there are no forces performing in the horizontal
direction, this means that there is no acceleration in the horizontal direction
and consequently the horizontal velocity is persistent.
Velocity = distance/ time.
Horizontal velocity is therefore horizontal distance/time = 0.61 m/0.39s = 1.56 m/s.
The autographed baseball rolled off at 1.564 m/s before it fell off.
Given the following data:
To find how fast the autographed baseball was rolling on the desk before it fell off:
First of all, we would use the second equation of motion to determine the time it took:
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
Substituting the given values into the formula, we have;
[tex]0.76 = 0(t) + \frac{1}{2} (9.81)t^2\\\\0.76 = 4.905t^2\\\\t^2 = \frac{0.76}{4.905}\\\\t^2 = 0.155\\\\t = \sqrt{0.155}[/tex]
Time, t = 0.39 seconds.
Next, we would determine the horizontal speed:
[tex]Horizontal \; speed = \frac{Horizontal \; distance}{Time}\\\\Horizontal \; speed = \frac{0.61}{0.39}[/tex]
Horizontal speed = 1.564 m/s
Therefore, the autographed baseball rolled off at 1.564 m/s before it fell off.
Read more: https://brainly.com/question/8898885
