Calculating Launch Speed for a Catapult with Drag Force

  • Thread starter Thread starter juju1
  • Start date Start date
  • Tags Tags
    Physics
Click For Summary
The discussion revolves around calculating the launch speed of a tennis ball from a catapult, considering a spring constant of 101 N/m and a stretch of 0.477 m. The students initially struggle with the correct application of the energy equations, particularly in distinguishing between the variables for height and distance. After some corrections, the calculated launch speed is approximately 19.7 m/s, factoring in significant figures. The conversation highlights the importance of accurately using given values and maintaining precision in calculations. Ultimately, the students arrive at a consensus on the correct launch speed.
juju1

Homework Statement



Two students, Joe and Jane Fisycks, make a catapult. They use a spring with a spring constant of 101 N/m to launch a tennis ball. When the catapult is loaded, the spring is stretched 0.477 m. This is also the distance the tennis ball travels as it is being fired by the catapult. When loaded, the tennis ball is at ground level. When it leaves the catapult, it is 0.567 m above ground level. A drag force D acts on the tennis ball as it is being fired. The tennis ball's mass is 0.0579 kg.

(e) If there were no drag, what launch speed would the tennis ball have? (find v_1)

Homework Equations

The Attempt at a Solution


so i found the equations when it is ready to fire and after it fires
.5kx^2 and (.5mv_1^2)+(mgy_1)+Dx

i put these two equations together to solve for v_1, yet I still am not getting a right answer.

my v_1 equation is sqrt( (.5kx^2 - mgy_1) / (.5m) )

i plugged in 101N/m for k
0.477 m for y_1
0.0578 for m
9.8 for g
0.567 for x

what am i doing wrong?

 
Physics news on Phys.org
Looks like you might have switched the values of x and y1.
 
so 19.659 m/s?
 
juju1 said:
so 19.659 m/s?
I believe that's right. Note that the mass is given to be 0.0579 kg, but you later wrote 0.578 kg. (Not a very significant difference.)

How many significant figures should the answer have?
 
3 sigfigs! So rather: 19.7 m/s
 
  • Like
Likes TSny
juju1 said:
3 sigfigs! So rather: 19.7 m/s
Yes. That's essentially what I got, too. (19.6 m/s)
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
2
Views
2K
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
5
Views
2K
Replies
4
Views
6K
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
5K