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Sagot :
Answer:
Max
Step-by-step explanation:
the function given that h = -16t³+64t+10
the general function Equation:
h = at²+bt+c
=> t at the highest point is defined by t = -b/2a
= -64/2(-16) = -64/-32 = 2 second
the total times that the ball hits the ground
= 2× 2 seconds = 4 seconds.
so, Max is right
Answer:
Solution given:
Let h=0=when the ball hit the ground,the height=0.
h= -16t^2 + 64t + 10.
0=16t²-64t-10
8t²-32t-5=0
∆=[tex] \frac{32±\sqrt{32²+4*5*8}}{2*8}=\frac{8±\sqrt{74}}{4}[/tex]
taking positive
t1=[tex] \frac{8+\sqrt{74}}{4}=4.15seconds[/tex]
t2=[tex] \frac{8-\sqrt{74}}{4}(t2<O)(neglected)[/tex]
So
t=4.15seconds closer to 4seconds.So
Max is closer.
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