Free Fall Problem: How to Calculate Time in Air for a Thrown Ball?

  • Thread starter Thread starter VoNi
  • Start date Start date
  • Tags Tags
    Fall Free fall
AI Thread Summary
To solve the free fall problem of a ball thrown upward with an initial velocity of 10 m/s from a height of 2.3 m, the correct equation is delta-y = (vi)t - 1/2gt². The user initially set up the quadratic equation correctly but made errors in applying the quadratic formula. The coefficients in the formula should be adjusted to reflect the correct signs, particularly using -10 instead of 10 for the velocity term. After correcting these mistakes, the user should arrive at the correct time in the air. Proper application of the quadratic formula is essential for finding the accurate solution.
VoNi
Messages
3
Reaction score
0
Free fall problem, Help!?

Homework Statement



Hi, I am having trouble figuring out this problem:

A ball is thrown directly upward with an initial velocity of 10 m/s. If the ball starts at an initial height of 2.3 m, how long is the ball in the air? Ignore air drag.

Homework Equations



The equation that I used to solve this was:

delta-y= (vi)t-1/2gt²

The Attempt at a Solution



I plugged in my numbers so i got:

-2.3= 10t-1/2(9.8)t²
-2.3=10t-(4.9)t²
4.9t²-10t-2.3=0

then i did the pythagorean theorem(but I don't really know how to type it on the computer.. but i will try)

-10 ± √((-10²)-4(4.9)(-2.3))/ 2(4.9)

-10 ± √(100-45.08)/2(4.9)

-10 ± √(54.92)/ 9.8

(-10 ± 7.41)/ 9.8

1.77 s or -1.77 s


So I put the answer 1.77 s and it was wrong. i put 1.8s and it was wrong. I used 1.02 and it was wrong. I used 2.04 and it was wrong. I have no idea what i am doing wrong and I no longer have any more submissions for my answer (which means i oficially have a zero for that question). But i really need to know how to solve this problem! please help me!

thanks!
 
Physics news on Phys.org


You just made a couple mistakes with your sign:

You have
4.9t²-10t-2.3=0 which is correct but when you use the quadratic formula you have (it's not the Pythagorean btw)

-10 ± √((-10²)-4(4.9)(-2.3))/ 2(4.9)

but it should be

(10 +/- sqrt[(-10)^2 - 4(4.9)(-2.3)])/(2(4.9))

which is

((10 +/- sqrt[100 + 4(4.9)(2.3)])/(2(4.9))


that should give you the correct answer
 


a=4.9
b=-10 (not 10!)
c=-2.3
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top