Impulse question: throwing a rock

1. Dec 10, 2013

Morange

Hello,
1. The problem statement, all variables and given/known data
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 travelled 1.20 m. What is the average force that you exert on the ball while it is in your hand?

2. Relevant equations
$\vec{F}$=$\frac{∆P}{∆t}$

3. 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

Last edited: Dec 10, 2013
2. Dec 10, 2013

Staff: Mentor

Gravity will give an additional ∆P you can include in your calculation.

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).

3. Dec 10, 2013

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?

4. Dec 10, 2013

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?

5. Dec 11, 2013

Staff: Mentor

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.