# Free Falling Object question

by veloix
Tags: falling, free, object
 P: 46 [b]1. The problem statement, all variables and given/known data[/ A ball is thrown directly downward, with an initial speed of 7.15 m/s, from a height of 31.0 m. After what time interval does the ball strike the ground? 2. Relevant equations All the kinematics equations dealing with velocity and height and accelration. need to solve for time. 3. The attempt at a solution I tried this one couple of times. The first thing I did was use the equation Yf=Yi+Viyt+1/2(a)t^2. I chose the top to be postive. 0= 31.0m+ 7.15 m/s(t)+1/2(-9.80)t^2 I saw that this look like a qudratic equation so i work it out ax+bx+cx -7.15 m/s +/-[(-7.15-4(-4.90)(31.0)/2(-4.90)]^1/2 it came out to be -7.15+7.53 = 0.38 -7.15-7.53= -14.68 i think im not doing something right, need some help with this one.
 P: 1,877 I didn't personally check it, but it seems fine. The ball would hit the ground within a couple seconds or so (approx sqrt(6)) if you just dropped it, so if you throw it down it will be even quicker, like the half a second you got.
 HW Helper P: 4,125 Viy = -7.15m/s since it is thrown downward.
P: 46

## Free Falling Object question

oh ok i going compute this out then ty.
 P: 46 hmm its still not comeing out right i check my work but i keep getting the wrong anwser. I dont know what else i can do.
HW Helper
P: 4,125
 Quote by veloix hmm its still not comeing out right i check my work but i keep getting the wrong anwser. I dont know what else i can do.
What time do you get?
 P: 46 when i caluclated everything t came out to be 15.34 if I chose the positve and -1.04 with the negative.
HW Helper
P: 4,125
 Quote by veloix when i caluclated everything t came out to be 15.34 if I chose the positve and -1.04 with the negative.
 P: 46 0= 31.0m + (-7.15 m/s)t + 1/2(-9.80 m/s^2)t^2 X= -(-7.15)+/-[-7.15^2-4(-4.90)(31.0)/2(-4.90)]^1/2 X=-(-7.15)+/-[-51.12-(-608)/-9.80]^1/2 X=-(-7.15)+/-[557/-9.80]^2 X=-(-7.15)+/-[-56.8]^1/2 hmm i made a mistake with my signs man thse quads are tough hehe. i guess the solutions i got before dont reall exsit because of the negative rooot. hmmmm
HW Helper
P: 4,125
 Quote by veloix 0= 31.0m + (-7.15 m/s)t + 1/2(-9.80 m/s^2)t^2 X= -(-7.15)+/-[-7.15^2-4(-4.90)(31.0)/2(-4.90)]^1/2 X=-(-7.15)+/-[-51.12-(-608)/-9.80]^1/2 X=-(-7.15)+/-[557/-9.80]^2 X=-(-7.15)+/-[-56.8]^1/2 hmm i made a mistake with my signs man thse quads are tough hehe. i guess the solutions i got before dont reall exsit because of the negative rooot. hmmmm
Hmmm... I don't think you're applying the quadratic equation properly. the /2a part doesn't go inside the square root... and you should have (-7.15)^2 = 51.12 not -7.15^2=-51.12
 P: 46 wow i was messing that up all kinds of ways i finally got the right anwser. thanks alot for your help you save me on endless path hehe. the right anwser was 1.89s using the positive side.
HW Helper
P: 4,125
 Quote by veloix wow i was messing that up all kinds of ways i finally got the right anwser. thanks alot for your help you save me on endless path hehe. the right anwser was 1.89s using the positive side.
cool! no prob!
 P: 15 sometimes it helps to reduce error and make life easier to plug in the x = vi t + 1/2 at^2 formula into the quadratic equation first and simplify before plugging in the actual values... this approach makes life easier for some ppl

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