So my exam is in 2 days and I am having so much trouble. I am a new member, thanks for taking the time to view this question.
If an object is thrown downward at 26 m/s from the top of a building, 102m high, the equation relating to displacement (meters) and time (seconds) is:
s = 1/2 gt^2 + 26t + 102
Find the time for an object to reach the ground from the top. Use g=-9.81
So I have come up with: (I am new to this; sq = square root symbol, sorry)
-26 +/- sq (26)^2 - 4(-4.905)(102) / -9.81
My problem occurs when my two answers, rougly, equal:
I am confused as to why I have 2 negative times. One should be positive and one negative, correct?
The displacement equation under constant acceleration is:
where is the acceleration, the initial velocity and the initial displacement.
In this problem , .
The minus signs are because we have chosen positive to indicate upwards and ground level to be