Answer:
c. o.5seconds
Step-by-step explanation:
y=-16t^2 +4t+2
=-2(8t^2 -2t-1)
=-2(8t^2 -4t+2t-1)
as the object hits the ground y =0
0 =-2(4t+1)(2t-1)
(4t+1)(2t-1)=0
t=-1/4 meangless
t=1/2
=0.5seconds
Answer:
If y is the height of the ball, then y = 0 when the ball is on the ground. Plug 0 in for y and use the quadratic equation to find the solutions:
0 equals short dash 16 t squared plus 4 t plus 2
t equals fraction numerator short dash 4 plus-or-minus square root of 4 squared minus 4 left parenthesis short dash 16 right parenthesis left parenthesis 2 right parenthesis end root over denominator 2 left parenthesis short dash 16 right parenthesis end fraction
t equals fraction numerator short dash 4 plus-or-minus square root of 16 minus open parentheses short dash 128 close parentheses end root over denominator short dash 32 end fraction
t equals fraction numerator short dash 4 plus-or-minus square root of 144 over denominator short dash 32 end fraction
t equals fraction numerator short dash 4 plus-or-minus 12 over denominator short dash 32 end fraction
t equals fraction numerator short dash 4 minus 12 over denominator short dash 32 end fraction space o r space fraction numerator short dash 4 plus 12 over denominator short dash 32 end fraction space
t equals 0.5 space O R space short dash 0.25
Since we are looking for a distance, only the positive answer of 0.5 is appropriate.
so, C. 0.5 seconds
Step-by-step explanation:
