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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