trinityclarkson
contestada

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 8 ft with an initial velocity of 28 ft/s.

After how many seconds does the ball hit the ground?

Respuesta :

WeatheringWithYou
It hits the ground after two seconds. 
apologiabiology
initial height is h0 which is 8
v0 is initila velocity which is 28

h(x)=-16x^2+28x+8 is equation
when it hits the ground, h(x)=0
so
solve
0=-16x^2+28x+8
hmm, we can factor out the -4
0=-4(4x^2-7x-2)
use ac method factor
what 4 times -2=-8
what 2 numbers multiply to get -8 and add to get -7?
-8 and 1
0=-4(4x^2-8x+1x-2)
0=-4((4x^2-8x)+(1x-2))
0=-4(4x(x-2)+1(x-2))
0=-4(4x+1)(x-2)
set to zero
4x+1=0
4x=-1
x=-1/4
false, we can't start before x=0

x-2=0
x=2

hits ground at x=2 or after 2 seconds

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