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The height, in feet, after x seconds of an object launched straight up can be found by the function h(x)=−16x2+v0x+h0, where v0 is the initial velocity of the object and h0 is the initial height.

A ball is kicked straight up into the air from a height of 12 ft with an initial velocity of 44 ft/s. After how many seconds does the ball hit the ground?

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___ s

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HomertheGenius
The function is:
h ( x ) = - 16 x² + vo x + h o
v o = 44 ft / s ( initial velocity )
h o = 12 ft ( initial height )
- 16 x² + 44 x + 12 = 0 / : 4  ( we will divide both sides by 4 )
- 4 x² + 11 x + 3 = 0
- 4 x² + 12 x - x + 3 = 0
- 4 x ( x - 3 ) - ( x - 3 ) = 0
( x - 3 ) ( - 4 x - 1 ) = 0
x - 3 = 0 ,   x = 3 s
or:
- 4 x - 1 = 0 ,  - 4 x = 1,  x = - 1/4 ( false ).
Answer:
3 seconds.

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