A basketball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?

0.7/0.25=2.8m per second

Answered by RhysH | 2024-06-10

The ball's speed is found to be approximately 0.663 m/s.
To determine how fast the basketball was rolling as it left the desk, we need to calculate its horizontal velocity. Since we are ignoring air resistance and assuming no horizontal forces are acting on the ball after it leaves the desk, the horizontal velocity remains constant.
We know the height of the desk (0.70 m) and the horizontal distance to where the ball strikes the floor (0.25 m). The time it takes for the ball to fall can be calculated using the equation for free-fall motion in the y-direction: h = 1/2gt², where h is the height, g is the acceleration due to gravity (approximately 9.81 m/s^2), and t is the time in seconds.
Step 1:
Solve for time (t):
0.70 m = 1/2(9.81 m/s^2)t^2
t² = (2 / 9.81 m/s^2)(0.70 m)
t = /((2 / 9.81 m/s^2)(0.70 m))
t /= sqrt(2 / 9.81 m/s^2)(0.70 m)) approx 0.377 seconds
Step 2:
Calculate horizontal velocity (v):
0.25 m = v(0.377 s)
v = 0.25 m / 0.377 s = approx 0.663 m/s

Answered by SanaPar | 2024-06-24

The basketball was rolling off the desk at approximately 0.663 m/s when it left the surface. This calculation was derived from the time it took to fall 0.70 m and the horizontal distance traveled. By using free fall equations and projectile motion, we found the initial speed before it hit the ground.
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Answered by SanaPar | 2024-10-15