Impulse question: throwing a rock

[Morange]

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?

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

 

[mfb]

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).

 

[Rolen]

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?

 

[Morange]

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?

 

[mfb]

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.