math
posted by cris
An object is thrown or fired straight upwards at an initial speed of v_0 ft⁄s will reach height h feet after t seconds, where h and t are related to the formula
h=16t^2+v_0 t
Suppose the object is fired straight upwards with an initial speed of 800ft⁄s, What is the initial velocity?
How does it change the equation h=16t^2+v_0 t?
What is the initial position of the object?
When does the object fall back to the ground?
When does the object reach a height of 6400ft?
When does the object reach a height of 2mi?
How high is the highest point the ball reaches?
Suppose the object is dropped from a height of 288ft, what is v_0?
The equation becomes h=16t^2+h_0 after (g) Why?
Write an equation which includes 288ft

Steve
initial velocity is 800 ft/s upward
it does not change the equation.
initial position is 0, since it was fired from the ground (height=0)
It falls back to the ground when h=0. So, solve for t in 800t16t^2 = 0
Solve for t in 80016t^2 = 6400
(why are there two solutions?)
Same as above, but you need to convert 2 miles to feet.
max height at the vertex of the parabola, when t is midway between the roots of the equation.
if dropped, v_0 is zero.
with v_0 = 0, the term vanishes, and we have an initial height, rather than initial speed.
... 
cris
thank you Steve for your help
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