# Free Falling Object question

1. Homework Statement [/
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?

## Homework Equations

All the kinematics equations dealing with velocity and height and accelration. need to solve for time.

## 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.

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.

learningphysics
Homework Helper
Viy = -7.15m/s since it is thrown downward.

oh ok i going compute this out then ty.

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.

learningphysics
Homework Helper
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?

when i caluclated everything t came out to be 15.34 if I chose the positve and -1.04 with the negative.

learningphysics
Homework Helper
when i caluclated everything t came out to be 15.34 if I chose the positve and -1.04 with the negative.

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

learningphysics
Homework Helper
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

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.

learningphysics
Homework Helper
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!

http://img215.imageshack.us/img215/5994/freefallgn3.jpg [Broken]

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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