A baseball rolls off a 0.70 m high desk and strikes the floor 0.25m away from the base of the desk. how fast was the ball rolling? please help… I don’t understand this :
The Correct Answer and Explanation is:
To find how fast the baseball was rolling when it left the desk, we can treat this as a projectile motion problem. Here’s how to solve it:
Correct Answer:
The ball was rolling at 0.67 m/s when it left the desk.
Step-by-Step Explanation
When a ball rolls off a desk, it follows a curved path downward because of gravity. The motion has two parts: horizontal and vertical. Since there’s no air resistance and the ball was not pushed downward, it only has initial velocity in the horizontal direction. Gravity pulls it down vertically.
Let’s start by analyzing the vertical motion. The desk is 0.70 meters tall, and the only force acting on the ball in this direction is gravity. We can use the equation for free fall to find the time it takes to hit the ground:
h=12gt2h = \frac{1}{2}gt^2h=21gt2
Where:
- h=0.70h = 0.70h=0.70 meters
- g=9.8 m/s2g = 9.8 \, \text{m/s}^2g=9.8m/s2
- ttt is time in seconds
Rearranging to solve for ttt:
t=2hg=2×0.709.8≈0.143≈0.378 secondst = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.70}{9.8}} \approx \sqrt{0.143} \approx 0.378 \, \text{seconds}t=g2h=9.82×0.70≈0.143≈0.378seconds
Now use this time to calculate the horizontal velocity. The ball travels 0.25 meters horizontally during this time. Since horizontal velocity stays constant (no horizontal acceleration), we use:
v=dt=0.250.378≈0.661 m/sv = \frac{d}{t} = \frac{0.25}{0.378} \approx 0.661 \, \text{m/s}v=td=0.3780.25≈0.661m/s
Rounding to two decimal places, the ball was rolling at 0.67 m/s.
This kind of problem is common in physics. It shows how vertical and horizontal motions are independent. The key idea is that the vertical fall tells us how long the ball was in the air, and from that time, we figure out how fast it must have been moving horizontally to cover the distance.
