Question

1. a 68.0 kg baseball player jumps into the air, and throws a 145 g baseball directly east. he throws the baseball with a force of 4.40 n. what is the resulting acceleration of the player? 0.064 m/s² west 0.064 m/s² east 0.030 m/s² west 0.030 m/s² east

Explanation:

Step1: Apply Newton's third law

According to Newton's third law, the force exerted on the player is equal in magnitude and opposite in direction to the force exerted on the baseball. So \(F = 4.40\ N\) (west, opposite to the direction of the force on the baseball).

Step2: Use Newton's second law \(F = ma\)

We know \(m = 68.0\ kg\) (mass of the player) and \(F = 4.40\ N\). Rearranging \(a=\frac{F}{m}\).
Substitute the values: \(a=\frac{4.40\ N}{68.0\ kg}\approx0.0647\ m/s^{2}\approx0.064\ m/s^{2}\) (direction is west as per Newton's third - law).

Answer:

0.064 m/s² west