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Elementary algebra help!!!?
A water rocket is launched upward with an initial velocity of 48ft/sec. Its height h, in feet, after t seconds is given by h=48t - 16t^2. Determine the height of the rocket 2.5 sec after it has been launched.
I need the procedure please, explainn me!!
7 Answers
- detektibgapoLv 51 decade ago
h = 48t - 16t^2
Since t = 2.5 sec, I guess you have just to plug it in the equation.
h = 48 (2.5) - 16 (2.5)^2
h = 120 - 16(6.25)
h = 120 - 100
h = 20 feet ===> the answer
- Wyk123Lv 41 decade ago
just substitute 2.5 sec. to the given equation:
h=48*2.5+16*(2.5)^2
h= 120 +100
h=320 ft
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- Anonymous1 decade ago
Just substitute 2.5 for t every time t occurs.
h is height.
h= 48t - 16t^2
h= 48(2.5) - 16(2.5)(2.5) now calculate it.
- 1 decade ago
Just plug in 2.5 seconds for t (since t is time here) so
48(t) -16(t)^2=h
this is basically using the energy equation (assuming no non-conservative forces)