Question 17128: You fire a rifle straight up. Your bullet leaves your fun at a velocity of 960 feet per second. Ignore air resistance. Consider the mussle of you gun to be at height zero. The bullet will reach its maxium altitude of __ feet after __ seconds.
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! I believe the formula for this is
 , where y is the number of feet above the ground t seconds after firing, assuming gravitational constant of -32 ft/sec^2.
Soooo,
Where, y=0 are the two points where the bullet is at ground level.
Factor the common factors of 16t, and while you are at it, why not take out a -16t. That's even better.
t= 0 and t=60 seconds.
The first value t=0 represents the time when the gun was fired, and the bullet left the gun. The second time t=60 seconds represents the time when the bullet came back down to ground level.
By symmetry, the bullet spends half the time going up, and half the time coming back down. So the bullet reaches its maximum altitude at t= 30 seconds.
Maximum height attained is
 , where t= 30 seconds
 feet
Check with a calculator:
(Notice that on the y axis of this graph, the ten thousands digit is not displayed. Sorry about that. Still a pretty nice graph if you will correct that minor error by algebra.com graphing.)
R^2 at SCC
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