Posted by cris on Sunday, May 25, 2014 at 3:09pm.
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^2v_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^2v_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

math  MathMate, Sunday, May 25, 2014 at 3:17pm
You may want to check the equation to see if there is a typo. I expect it to read:
h=16t²+v_0 t
since the object is thrown straight upwards, while gravity (downwards) is indicated negative.
If it is fired straight upwards at an initial speed of 800 ft/s, the initial velocity is +800 ft/s, i.e. upwards.

math  cris, Tuesday, May 27, 2014 at 2:27pm
yes, you're right. the equation is h=16t^2+v_0t

math  mike wallon, Tuesday, February 21, 2017 at 11:19am
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