Question
A model airplane is shot up from a platform 1 foot above the ground with an initial upward velocity of 56 feet per second. The height of the airplane above ground after t seconds is given by the equation h=-16t^+56t+1, where h is the height of the airplane in feet and t is the time in seconds after it is launched. Approximately how long does it take the airplane to reach its maximum height?
A.
0.3 seconds
B.
1.8 seconds
C.
3.5 seconds
D.
6.9 seconds
A.
0.3 seconds
B.
1.8 seconds
C.
3.5 seconds
D.
6.9 seconds
Answers
feet units drive me crazy but all right.
16 t^2 - 56 t - 1 = -h
16 t^2 - 56 t = - h + 1
t^2 - 3.5 t = - h/16 + 1/16
t^2 - 3.5 t + (3.5/2)^2 = -h/16 + 1/16 + 12.25/4
(t - 3.5/2)^2 = ......
vertex (top) when t = 3.5/2 = 1.75 seconds
if they ask how high, keep going with completing the square.
16 t^2 - 56 t - 1 = -h
16 t^2 - 56 t = - h + 1
t^2 - 3.5 t = - h/16 + 1/16
t^2 - 3.5 t + (3.5/2)^2 = -h/16 + 1/16 + 12.25/4
(t - 3.5/2)^2 = ......
vertex (top) when t = 3.5/2 = 1.75 seconds
if they ask how high, keep going with completing the square.
By the way, if you happen to know Physics:
v = Vi - 32 t
at top t = 0
0 = 56 - 32 t
t = 1.75 seconds
v = Vi - 32 t
at top t = 0
0 = 56 - 32 t
t = 1.75 seconds
By the way, if you happen to know Physics:
v = Vi - 32 t
at top v = 0
0 = 56 - 32 t
t = 1.75 seconds
32 ft/sec^2 = acceleration of gravity in these obsolete units
v = Vi - 32 t
at top v = 0
0 = 56 - 32 t
t = 1.75 seconds
32 ft/sec^2 = acceleration of gravity in these obsolete units
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