Question

You throw a 5.25 ounce (0.15 kg) baseball at 105 miles per hour (47 m/s) towards the batter. If the time it took the pitcher to throw the baseball were 0.20 seconds, how much force did the pitcher apply to the baseball?

Answer 1

The pitcher applied a force of 35.25 newtons to the baseball, using the mass of the baseball, the final velocity, and the time interval to calculate acceleration, and then force via Newton's second law.

Step-by-step explanation:

To determine the force the pitcher applied to the baseball, we can use the formula derived from Newton's second law of motion, F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration. The acceleration can be found using the formula a = Δv/Δt, where Δv is the change in velocity and Δt is the time over which the change occurred.

Given that the baseball's mass (m) is 0.15 kg and it was thrown at a speed (v) of 47 m/s, and the time (t) taken to throw the baseball was 0.20 seconds, we can first calculate the acceleration.

As the baseball starts from rest, its initial velocity (u) is 0 m/s, so the change in velocity (Δv) is simply 47 m/s - 0 m/s = 47 m/s. Therefore, the acceleration (a) is 47 m/s divided by 0.20 s, which equals 235 m/s².

Now we can calculate the force:

F = ma
F = 0.15 kg × 235 m/s²
F = 35.25 N

Hence, the pitcher applied a force of 35.25 newtons to the baseball.