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babbagee
Oct2-05, 05:07 PM
An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point a distance 50.0 m below its starting point at a time 5.00 s after it leaves the thrower's hand. Air resistance may be ignored.

I tried alot of methods but I keep getting it wrong. I am doing this online so it tells me weather the answer is right or wrong. I tried finding the time it takes to travel the 50m and then subtracting that from the total time to see how long it was in the air from the origin to the time it reaches the origin again. But that did not seem to work. Can someone point me in the right direction. I dont want the answer just give me some hints.

Thanks

Integral
Oct2-05, 05:23 PM
You did not state the complete problem. What are you looking for?

babbagee
Oct2-05, 05:24 PM
sorry about that,

a.) What is the initial speed of the egg?

b.) How high does it rise above its starting point?

the second one is easy once i find the intial speed. So I think i only need help with a.

thanks

Kazza_765
Oct2-05, 05:26 PM
You will need to use one of the equations of motions, but you haven't specified what the question is actually asking you. You are probably having problems here because you are not defining which direction is positive. So pick either up or down to be the positive direction. For example, if up is positive and down is negative, then initial velocity is positive, accelleration is negative, total displacement is -50m. As long as you are consistent with your sign convention, you should be able to just put all the values into one of the formulas and get an answer.

babbagee
Oct2-05, 05:35 PM
It asks for the intial velocity, and I am not having any trouble with my signs. I been using up as positive and down as negative.

HallsofIvy
Oct2-05, 06:20 PM
Okay so h(t)= -4.9t2+ vt where "v" is the (unknown) initial velocity (yes, + up, -down) and we are taking the height of the roof to be "0". You know that when t= 5, h(5)= -50. Put those into the equation and solve for v.

babbagee
Oct2-05, 08:36 PM
Hey thanks,

I was thinking of spliting the problem into two parts, but i guess the eqations take that into consideration, right.