When Will the Thrown Ball Reach the Ground?

[veloix]

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 positive.
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 I am not doing something right, need some help with this one.

 

[Mindscrape]

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]

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

 

[veloix]

oh ok i going compute this out then ty.

 

[veloix]

hmm its still not comeing out right i check my work but i keep getting the wrong anwser. I don't know what else i can do.

 

[learningphysics]

hmm its still not comeing out right i check my work but i keep getting the wrong anwser. I don't know what else i can do.

What time do you get?

 

[veloix]

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

 

[learningphysics]

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

can you show your calculations?

 

[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 don't reall exsit because of the negative rooot. hmmmm

 

[learningphysics]

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 don't 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

 

[veloix]

wow i was messing that up all kinds of ways i finally got the right anwser. thanks a lot for your help you save me on endless path hehe. the right anwser was 1.89s using the positive side.

 

[learningphysics]

wow i was messing that up all kinds of ways i finally got the right anwser. thanks a lot for your help you save me on endless path hehe. the right anwser was 1.89s using the positive side.

cool! no prob!

 

[circuscircus]

http://img215.imageshack.us/img215/5994/freefallgn3.jpg

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