A 0.15 kg baseball moving at +26 m/s is slowed to a stop by a catcher who exerts a constant force of -390 N. How long does it take this force to stop the ball? How far does the ball travel before stopping?

plz plz plz help!!

.01s

Well, well, well, looks like we have a real-life game of catch here! Let me crunch some numbers and get back to you with the answer.

To find the time it takes to stop the ball, we can use the equation:

Force equals mass times acceleration (F = ma)

Given that the force is -390 N and the mass is 0.15 kg, we can rearrange the equation to solve for acceleration:

Acceleration = Force / Mass = -390 N / 0.15 kg

And guess what! Negative acceleration means that the ball is slowing down.

So, the acceleration of the ball is about -2600 m/s². Now let's find the time it takes for the ball to stop:

Using the equation of motion:

Final velocity = Initial velocity + (acceleration * time)

Since the final velocity of the ball is 0 m/s (it stops), and the initial velocity is +26 m/s, we can plug in the values:

0 m/s = 26 m/s + (-2600 m/s² * time)

Now let's solve for time:

26 m/s = 2600 m/s² * time
time = 26 m/s / 2600 m/s²

And after some mathematical magic, we'll find that it takes a whopping 0.01 second for the ball to come to a halt. Quick ball!

Now, to find how far the ball travels before stopping, we can use another equation of motion:

Distance = Initial velocity * time + (1/2) * acceleration * time²

Plugging in the values:

Distance = 26 m/s * 0.01 s + (1/2) * (-2600 m/s²) * (0.01 s)²

After some number crunching, the ball would travel approximately 0.13 meters before stopping.

So, there you have it! It takes 0.01 second for the catcher to stop the ball and it travels about 0.13 meters before coming to a halt. Just enough time for the catcher to show off their skills and prevent any home runs!

To find the time it takes for the force to stop the ball and the distance the ball travels before stopping, we can use the equations of motion.

First, let's find the deceleration (negative acceleration) of the ball using Newton's second law:

F = m * a

Where:
F = force applied by the catcher = -390 N (negative sign indicates opposite direction)
m = mass of the baseball = 0.15 kg

Rearranging the equation, we have:

a = F / m

a = -390 N / 0.15 kg
a = -2600 m/s²

Next, we can use the equation of motion to calculate the time it takes for the ball to stop:

v = u + a * t

Where:
v = final velocity = 0 m/s (since the ball stops)
u = initial velocity = +26 m/s (positive sign indicates direction)
a = deceleration = -2600 m/s² (negative sign indicates opposite direction)
t = time

Rearranging the equation, we have:

0 = 26 m/s + (-2600 m/s²) * t

Solving for t, we get:

26 m/s = 2600 m/s² * t
t = 26 m/s / 2600 m/s²
t ≈ 0.01 s

Therefore, the time it takes for the force to stop the ball is approximately 0.01 seconds.

To find the distance traveled by the ball before stopping, we can use the equation of motion:

s = u * t + (1/2) * a * t²

Where:
s = distance traveled
u = initial velocity = 26 m/s
t = time taken = 0.01 s
a = deceleration = -2600 m/s²

Plugging in the values, we get:

s = 26 m/s * 0.01 s + (1/2) * (-2600 m/s²) * (0.01 s)²
s = 0.26 m - 0.013 m
s ≈ 0.25 m

Therefore, the ball travels approximately 0.25 meters before stopping.

To find the time it takes for the force to stop the ball, 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 catcher and the acceleration is the change in velocity.

The force exerted by the catcher is -390 N, which means it is acting in the opposite direction of the ball's motion. Since we are given that the ball is slowing down, the acceleration will also be in the opposite direction of its initial velocity.

We can find the acceleration using the formula:
acceleration = (final velocity - initial velocity) / time

In this case, the final velocity is 0 m/s (since the ball stops), the initial velocity is +26 m/s, and we need to find the time. Rearranging the formula, we have:
time = (final velocity - initial velocity) / acceleration

Plugging in the values, we get:
time = (0 m/s - 26 m/s) / acceleration

To find the acceleration, we can use Newton's second law, which can be rearranged to:
acceleration = net force / mass

In this case, the net force is -390 N and the mass is 0.15 kg. Plugging in these values, we have:
acceleration = -390 N / 0.15 kg

Now, we can substitute this value into the time equation:
time = (0 m/s - 26 m/s) / (-390 N / 0.15 kg)

By simplifying the equation, we get:
time = -26 m/s / (-390 N / 0.15 kg)

To solve this equation, we divide the numerator by the denominator:
time = -26 m/s * (0.15 kg / -390 N)

Now, canceling out the appropriate units, we get:
time = (26 * 0.15) / 390 s

Calculating this expression, we find:
time ≈ 0.01 s

Therefore, it takes approximately 0.01 seconds for the catcher's force to stop the ball.

To find the distance the ball travels before stopping, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2

Since the final velocity is 0 m/s and we've already found the time and acceleration, we can plug in these values and simplify the equation. Calculating this expression, we find:
distance ≈ 0.01 m

Therefore, the ball travels approximately 0.01 meters before stopping.

Use the impulse-momentum wethod to answer this question. The impulse is the average force Fav, multipled bu the tme that it is applied (which is the time it takes to decelerate the ball).

|Fav| * T = 0.15 kg * 26 m/s

We only need the absolute value of the force, so never mind the minus sign in front of -390N.

T = .015*26/390 = ? seconds

The distance X travelled while stopping equals average velocity x Time
X = (1/2)V^2*M/Fav