Impulse question: throwing a rock

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The discussion centers on calculating the average force exerted on a 0.500 kg rock thrown underhand at a 40° angle. Participants clarify that to find the average force, one must consider both the horizontal and vertical components of acceleration, using kinematics equations. The calculations reveal that the net forces in both directions must account for gravity, leading to a total applied force of approximately 16.9 N. It is emphasized that the problem assumes constant acceleration, making the average force equal to the applied force. The conclusion highlights the need for clear problem statements in physics to avoid confusion.
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Hello,

Homework Statement


You throw a 0.500 kg rock underhand, as shown in the diagram. The rock starts at rest from the lowest point moving at an angle of 40° above the horizontal, and leaves your hand 0.300 s later having traveled 1.20 m. What is the average force that you exert on the ball while it is in your hand?
GRabT8m.png


Homework Equations


\vec{F}=\frac{∆P}{∆t}


The Attempt at a Solution


I tried to sub the values straight in but not sure now to incorporate the force of gravity. Also I used the average speed for ∆P

F = 0.500kg*(1.2m/0.300s)/0.300s = 6.67 N
 
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Gravity will give an additional ∆P you can include in your calculation.

Also I used the average speed for ∆P
That looks wrong.

I think you have to assume a uniform acceleration here (that should be given - without that, it is not possible to calculate the average force).
 
By average force the question means the minimum force for the object to leave your hand? I didn't quite get the final data, like, where's the object land? By the data that you show I'm not show how to calculate this. Are there more date to the problem?
 
This was all the data given. I think I have it:

I found the acceleration using the kinematics equation

∆d = t*V1 + 1/2*a*t^2

in which ∆d = 1.2, V1 = 0, and t = 0.300s

1.2m = 0.300s*(0 m/s) + 1/2a(0.300)^2
then the a turns out to be 26.666 m/s

Then I broke the acceleration into x,y components
a(x) = 26.6666*cos(40) = 20.427
a(y) = 26.6666*sin(40) = 17.141

Then I calculated the net force for each component using F = m*a, m = 0.500kg
Fnet(x) = F(x, applied)
= 20.427m/s^2 * 0.500 kg
= 10.21 N

Fnet(y) =
F(y, applied) - Fg
= 17.141 m/s^2 *0.500 kg
= 8.57 N

Since the vertical component of net force is the sum of gravity and the applied force I add the force of gravity to the vertical net force
8.57 N + mg = 8.57 N + (0.500)(9.8) = 13.47 N

Now that I have the horizontal and vertical applied forces I used pythagorean theorem to figure out the applied force
F(applied) = \sqrt{13.47^2+10.21^2}
= 16.9 N

Now I question I have is why is 16.9 N the average force AND the applied force? Is it due to it being constant force?
 
With the assumptions made in your derivation, it is a constant force, and the average of a constant value is always this constant value.

As I wrote before, you have to make an assumption about the acceleration process to get a result (--> bad problem statement), so a constant acceleration (and therefore a constant force) is the most reasonable choice.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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