SOLUTION: I would like to just say thank you for answering my question in advance. My question is: An object is thrown upward with a velocity of 45.0 m/s. When will it ne 25.0 m above

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Question 1011089: I would like to just say thank you for answering my question in advance.
My question is:
An object is thrown upward with a velocity of 45.0 m/s. When will it ne 25.0 m above its initial position.
The formula to use is: Vot + 1/2(gt^2)
Thank you.
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
H(t)=-4.9t^2+45t=25
-4.9t+45t-25=0=4.9t^2-45t+25
t=(1/9.8)(-45+/-sqrt (2025-122.5); sqrt term=39.18
t=(-1/9.8)(-5.82)=0.594 sec
t=(-1/9.8)(-84.18)=8.59 sec.
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graph%28300%2C200%2C-1%2C3%2C-10%2C50%2C-4.9x%5E2%2B45x%29
It is negative -4.9t^2 to take into account that gravity is pulling the object down while Vot is pushing the object up.