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.
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
3 answers
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
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
great job