A baseball catcher puts on an exhibition by catching a 0.150 kg ball dropped from a helicopter at a height of 100 meters above the catcher. If the catcher "gives" with the ball for a distance of 0.750 meters while catching it, what average force is exerted on the mitt by the ball? (g = 9.80 m/s^2)
a. 78N
b. 119N
c. 197N
d. 392N
The equation I have for average force is "Favg=delta (m*v) / delta time" but I don't think I can apply it directly to this situation.
How do you solve this problem?
You could use your equation, but would need one more step to be able to use the "give" distance X instead of the "delta time" of deceleration.
The average velocity of the ball while being decelerated is v/2. The "give" distance is thus
X = (v/2)*(delta t)
You will also need
v = sqrt(2gh) = 44.3 m/s
(neglecting air friction)
Favg = m*v/(2x/v) = (1/2)mv^2/X
= 197 N
Well, let's see if we can catch the solution to this problem!
To solve this, we can start by finding the initial potential energy of the ball when it's dropped from the helicopter. We can use the formula: PE = m * g * h, where m is the mass of the ball (0.150 kg), g is the acceleration due to gravity (9.80 m/s^2), and h is the height (100 meters).
PE = 0.150 kg * 9.80 m/s^2 * 100 m = 147 J
Now, let's find the final potential energy of the ball when it reaches the catcher's mitt. Since the ball is caught after giving for a distance of 0.750 meters, the height it drops to is 100 - 0.750 = 99.250 meters.
FE = 0.150 kg * 9.80 m/s^2 * 99.250 m = 145.78 J
Now, the difference in potential energy is the work done by the average force exerted on the mitt. So, we can subtract the final potential energy from the initial potential energy:
ΔPE = PE - FE = 147 J - 145.78 J = 1.22 J
But we need to convert this work into force. To do that, we can divide the work by the distance the ball gives:
ΔPE = Favg * d
1.22 J = Favg * 0.750 m
Now we can solve for Favg:
Favg = 1.22 J / 0.750 m = 1.6267... J/m
However, this is the force in joules per meter, and we need it in newtons. So, we divide by the distance in meters again to get the average force in newtons:
Favg = 1.6267... J/m / 0.750 m = 2.169 N
Hmm, it looks like none of the answer choices provided match our calculation. Maybe catching a ball dropped from a helicopter really is a unique experience!
To solve this problem, we can calculate the average force exerted on the mitt by the ball using the concept of work and energy.
First, let's determine the potential energy of the ball when it is dropped from a height of 100 meters. The potential energy is given by the equation:
Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
Substituting the given values:
PE = 0.150 kg * 9.80 m/s^2 * 100 m
PE = 147 J
Next, let's calculate the distance over which the catcher "gives" with the ball, which is 0.750 meters. This distance is subtracted from the initial height of 100 meters.
Distance (d) = 100 m - 0.750 m
Distance (d) = 99.25 m
Now, we can use the work-energy principle to find the average force exerted on the mitt.
Work done by gravity (W) = change in potential energy (ΔPE)
The work done by gravity is given by the equation:
W = force (F) * distance (d)
Rearranging the equation:
F = W / d
Substituting the values:
F = ΔPE / d
F = 147 J / 99.25 m
F ≈ 1.481 N
Therefore, the average force exerted on the mitt by the ball is approximately 1.481 N. None of the given answer choices match this value, which suggests a calculation error or a rounding difference.
Please double-check your calculations or review the question to ensure the correct values are provided.
To solve this problem, you need to break it down into smaller steps and use the appropriate equations. Here's how to approach it:
Step 1: Calculate the final velocity of the ball just before it reaches the catcher.
Start by determining the time it takes for the ball to fall from a height of 100 meters. You can use the equation for free fall motion:
h = (1/2) * g * t^2
Where:
h = height (100 meters)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearrange the equation to solve for time:
t = √(2h / g)
t = √(2 * 100 / 9.8) ≈ 4.52 seconds
Now, find the final velocity using the formula:
v = g * t
v ≈ 9.8 * 4.52 ≈ 44.34 m/s
Step 2: Calculate the initial velocity of the ball just before it reaches the catcher.
Since the ball is dropped from rest, its initial velocity is zero.
Step 3: Calculate the change in velocity.
The change in velocity is the difference between the final and initial velocities:
Δv = v - u
Δv = 44.34 - 0 ≈ 44.34 m/s
Step 4: Calculate the average force exerted on the mitt by the ball.
To find the average force, you can use Newton's second law of motion:
Favg = (m * Δv) / Δt
Where:
Favg = average force
m = mass of the ball (0.150 kg)
Δv = change in velocity (44.34 m/s)
Δt = time taken by the catcher to give with the ball (0.750 m)
Let's calculate Δt:
Δt = Δs / v
Δt = 0.750 / 44.34 ≈ 0.0169 seconds
Now, substitute the values into the equation:
Favg = (0.150 * 44.34) / 0.0169
Favg ≈ 0.3939 ≈ 392N
Therefore, the average force exerted on the mitt by the ball is approximately 392N.
Hence, the correct answer is (d) 392N.