When Reggie stepped up to the plate and hit a 0.150-kg fast ball traveling

at 36.0 m/s, the impact caused the ball to leave his bat with a velocity of
45.0 m/s in the opposite direction.
a. If the impact lasted for 0.002 s, what force did Reggie exert on the baseball?
b. Suppose the ball had instead travelled upwards at an angle of 30°. Find the force Reggie
exerted on the baseball in vector form. Then find the magnitude of the force Reggie
exerted.
I need help with B

a. V = Vo + a*t.

-45 = 36 + a*0.002, a = ?

F = M*a.

b. V = Vo + a*t.
45[30o] = 36 + a*0.002,
39 + 22.5i = 36 + a*0.002,
3 + 22.5i = a*0.002,
a = 1500 + 11,250i = 11,350 m/s^2[82.4o].

F = M*a = 0.15 * 11,350[82.4o] = 1,702N[82.4o].

Magnitude = 1,702 N.

I have question

To find the force Reggie exerted on the baseball when it travels upwards at an angle of 30°, we first need to break down the velocity of the ball into its x and y components.

Given:
Mass of the baseball (m) = 0.150 kg
Initial velocity of the ball (vi) = 36.0 m/s
Final velocity of the ball (vf) = -45.0 m/s (since it travels in the opposite direction)
Impact time (Δt) = 0.002 s
Angle of the ball's motion (θ) = 30°

a. Calculate the force exerted by Reggie on the baseball:

Let's use the impulse-momentum principle:

Impulse (J) = Change in momentum (Δp)
J = m * Δv

First, find the change in velocity:
Δv = vf - vi

Δv = -45.0 m/s - 36.0 m/s
Δv = -81.0 m/s

Now, calculate the impulse:
J = 0.150 kg * (-81.0 m/s)
J = -12.15 kg·m/s (negative sign indicates opposite direction)

Force (F) = Impulse / Time
F = J / Δt

F = -12.15 kg·m/s / 0.002 s
F = -6075 N

Therefore, Reggie exerted a force of -6075 N (in the opposite direction) on the baseball.

b. Calculate the force exerted by Reggie on the baseball in vector form and find its magnitude:

Given the upward angle of 30°, the x-component of the force would be negative and the y-component would be positive.

Force in x-direction:
Fx = F * cos(θ)
Fx = -6075 N * cos(30°)
Fx = -6075 N * 0.866
Fx = -5265.75 N

Force in y-direction:
Fy = F * sin(θ)
Fy = -6075 N * sin(30°)
Fy = -6075 N * 0.5
Fy = -3037.5 N

The force exerted by Reggie on the baseball in vector form is (-5265.75 N) * î + (-3037.5 N) * ĵ.

To find the magnitude of the force, use the Pythagorean theorem:
|F| = √(Fx^2 + Fy^2)

|F| = √((-5265.75 N)^2 + (-3037.5 N)^2)
|F| = √(27748006.56 N^2 + 9223878.75 N^2)
|F| = √(36971885.31 N^2)
|F| = 6079.72 N

Therefore, the magnitude of the force Reggie exerted on the baseball when it travels upwards at a 30° angle is approximately 6079.72 N.

To find the force Reggie exerted on the baseball when it travels upwards at an angle of 30°, we need to break down the problem into its x and y components.

First, let's find the x-component of the force. We know that the ball left Reggie's bat with a velocity of 45.0 m/s in the opposite direction. Since the ball is traveling upwards at an angle of 30°, we can use trigonometry to find the x-component of the velocity.

The x-component can be found using the equation:
Vx = V * cosθ
where Vx is the x-component of the velocity, V is the total velocity (45.0 m/s), and θ is the angle of 30°.

Plugging in these values, we have:
Vx = 45.0 m/s * cos(30°)
Vx = 45.0 m/s * 0.866 (approximating the cosine of 30°)

Calculating this, we find that Vx ≈ 39.0 m/s.

Since we know that force (F) is equal to the change in momentum (Δp) divided by the time (Δt), we can use this equation to find the x-component of the force:

F = Δp/ Δt
F = m(Vx - Vinitial)/ Δt

In this case, the initial velocity (Vinitial) of the baseball before Reggie hit it is 36.0 m/s, the mass (m) is 0.150 kg, and the impact lasts for 0.002 s.

Plugging in these values, we have:
F = 0.150 kg * (39.0 m/s - 36.0 m/s) / 0.002 s

Calculating this, we find that the x-component of the force is approximately 2,250 N.

Next, let's find the y-component of the force. We can use the same equation, but this time, we'll find the y-component of the velocity.

The y-component can be found using the equation:
Vy = V * sinθ
where Vy is the y-component of the velocity, V is the total velocity (45.0 m/s), and θ is the angle of 30°.

Plugging in these values, we have:
Vy = 45.0 m/s * sin(30°)
Vy = 45.0 m/s * 0.5 (approximating the sine of 30°)

Calculating this, we find that Vy ≈ 22.5 m/s.

Now, let's find the y-component of the force using the same equation:

F = Δp/ Δt
F = m(Vy - Vinitial)/ Δt

Plugging in the values, we have:
F = 0.150 kg * (22.5 m/s - 0 m/s) / 0.002 s

Calculating this, we find that the y-component of the force is approximately 1687.5 N.

To express the force Reggie exerted on the baseball in vector form, we combine the x and y components:
Force = (Fx, Fy)
Force = (2250 N, 1687.5 N)

To find the magnitude of the force Reggie exerted, we can use the Pythagorean theorem:
|Force| = √(Fx^2 + Fy^2)
|Force| = √(2250^2 + 1687.5^2)

Calculating this, we find that the magnitude of the force Reggie exerted is approximately 2822 N.