Posted by steve Alexander on Friday, October 24, 2008 at 7:24pm.
an object is fired upward at the end of the burn it has an upward velocity of 245m/s and is 14.7 m high
a) find the maximum height and when it is attained
b) when it reaches the ground
i know that i have to use the quadratic function s= 4.9t²+Vot+h
but for what i have the 4.9t² how can i use it and can help me with the problem please

calculus  bobpursley, Friday, October 24, 2008 at 7:30pm
For the max height, energy thinking
top PEnergy = intial PE + Initial Kenergy
mgh=mg*14.7 + 1/2 m 245^2
solve for h, the maximum height.
when is it attained?
h=14.5+245t 1/2 g t^2 for h max, solve for time t.
when it reaches the ground?
h=0 solve for time.

calculus  Reiny, Friday, October 24, 2008 at 7:40pm
The 4.9 is a fixed constant and deals with the force of gravity on earth.
Your equation would be
s = 4.9t^2 + 245t + 14.7
take the derivative set it equal to zero and solve for t
put that t back into the equation to find the maximum height
b) when it hits the ground, s = 0
so solve 0 = 4.9t^2 + 245t + 14.7 for t

calculus  Anonymous, Saturday, October 6, 2012 at 3:41am
great job
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