How Fast Was the Baseball Rolling Before It Fell?

[yankee]

An autographed baseball rolls off a 1.2m high desk & strikes the floor 0.37m away from the desk. The acceleration of gravity is 9.81m/s^2. How fast was it rolling on the desk before it fell off?

 

[kreil]

Analyze the problem vertically using the equation

$d=v_it+\frac{1}{2}at^2$

since vi is zero, solving for time yields:

$d=\frac{1}{2}at^2 \implies t=\sqrt{\frac{2d}{a}}$

Now, use this time to analyze the problem horizontally. It moved a total distance d in time t, so use the equation

$v_i=\frac{d}{t}$

and this will yield the balls initial horizontal velocity.

 

[kyle8921]

To calculate the initial velocity of the baseball before it fell off the desk, we can use the equation for projectile motion:

d = v0t + 1/2at^2

Where:
d = distance (0.37m)
v0 = initial velocity (unknown)
t = time (unknown)
a = acceleration due to gravity (9.81m/s^2)

We can also use the fact that the final velocity just before the ball hits the ground is 0, since it comes to a stop upon impact. Therefore, we can rewrite the equation as:

0 = v0t + 1/2at^2

Solving for t, we get:

t = √(2d/a)

Substituting the given values, we get:

t = √(2(0.37m)/(9.81m/s^2)) = 0.271s

Now, we can use this value of t in the original equation to solve for v0:

0.37m = v0(0.271s) + 1/2(9.81m/s^2)(0.271s)^2
0.37m = v0(0.271s) + 0.362m
v0 = (0.37m - 0.362m)/0.271s
v0 = 0.029m/s

Therefore, the baseball was rolling at a speed of 0.029m/s on the desk before it fell off.