The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 12.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 177 ms. The ball starts from rest.

(a) Through what distance does it move before its release?
(b) What are the magnitude and direction of the force the pitcher exerts on the ball?

I got part a) 1.062 but i cant get part b correct please help

a = change in velocity/time

a = (12-0)/ (177* 10^-3) = 67.8 m/s^2

distance = average velocity * time
= 6 * 177*10^-3 = 1.06 meters

F = m a
m = 2.28/9.81 = .232 kg
F = .232 * 67.8 = 15.7 horizontal force
there is also a vertical force up by the pitcher = 2.28 to keep it horizontal
so
F^2 = 2.28^2 + 15.7^2
angle up T
tan T = 2.28/15.7

To solve part b), we need to analyze the motion of the ball during the time interval before its release.

First, let's find the acceleration of the ball using the velocity and time given:
Given: initial velocity (u) = 0 m/s, final velocity (v) = 12.0 m/s, time (t) = 177 ms = 0.177 s.
Using the equation v = u + at, we can solve for acceleration (a):
12.0 m/s = 0 + a * 0.177 s
a = 12.0 m/s / 0.177 s
a ≈ 67.8 m/s²

Since the ball is moving horizontally, the only external horizontal force acting on it is the force exerted by the pitcher.

To find the magnitude of the force exerted by the pitcher, we can use Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration.
The mass of a baseball is approximately 0.145 kg.

F = 0.145 kg * 67.8 m/s²
F ≈ 9.82 N

Therefore, the magnitude of the force the pitcher exerts on the ball is approximately 9.82 N.

The direction of the force is horizontally, in the same direction as the motion of the ball.

So, the magnitude and direction of the force the pitcher exerts on the ball are 9.82 N horizontally.

To solve part (b), we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the force exerted by the pitcher on the ball.

We know that the only force acting on the ball horizontally is the force exerted by the pitcher, so we can equate the net force to the force exerted by the pitcher:

F_net = F_pitcher

We also know that the net force is equal to the product of the mass of the ball and its horizontal acceleration. However, in this scenario, the ball is not only subject to the horizontal gravitational force but also the horizontally applied force by the pitcher. Hence, we can write:

F_net = ma_horizontal

We can rearrange this equation to solve for the acceleration:

a_horizontal = F_net / m

To find the magnitude and direction of the force exerted by the pitcher, we need to find the horizontal acceleration of the ball. We can calculate the horizontal acceleration using the following formula:

a_horizontal = Δv / Δt

where Δv is the change in velocity and Δt is the time interval over which the acceleration occurs.

We know that the ball is thrown horizontally with an initial velocity of 0 m/s and a final velocity of 12.0 m/s. Thus, the change in velocity is:

Δv = 12.0 - 0 = 12.0 m/s

The time interval over which the acceleration occurs is given as 177 ms, which is equal to 0.177 s.

Substituting the values into the formula, we get:

a_horizontal = 12.0 m/s / 0.177 s ≈ 67.8 m/s^2

Now that we have the value of the horizontal acceleration, we can find the magnitude of the force exerted by the pitcher using Newton's second law:

F_net = ma_horizontal

Given that the mass of the ball is not mentioned in the question, we cannot directly calculate the magnitude of the force exerted by the pitcher. However, we can still determine the direction of the force. Since the ball is thrown horizontally, the force exerted by the pitcher must be in the same direction, i.e., horizontally.

Therefore, the force exerted by the pitcher on the ball has a magnitude that depends on the mass of the ball, but its direction is horizontal.

sin^-1(y/x)