Question

An autographed baseball rolls off of a 0.78 m

high desk and strikes the floor 0.25 m away
from the desk.
How fast was it rolling on the desk before
it fell off? The acceleration of gravity is
9.81 m/s
2
.
Answer in units of m/s

Answer 1

Step-by-step explanation:

We can use conservation of energy to solve this problem. At the top of the desk, the baseball has potential energy equal to mgh, where m is its mass, g is the acceleration due to gravity, and h is the height of the desk. At the bottom of the desk, all of this potential energy has been converted to kinetic energy, which is given by (1/2)mv^2, where v is the velocity of the baseball just before it hits the floor. We can set these two expressions equal to each other and solve for v:

mgh = (1/2)mv^2

We can simplify this expression by canceling out the mass:

gh = (1/2)v^2

Then, we can solve for v:

v = sqrt(2gh)

We know that h = 0.78 m and the acceleration due to gravity is g = 9.81 m/s^2, so we can substitute these values and calculate:

v = sqrt(2 × 9.81 m/s^2 × 0.78 m) = 3.41 m/s

Therefore, the baseball was rolling on the desk at a speed of 3.41 m/s before it fell off.