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Sagot :
Answer:
(a) 40 m
(b) 4 seconds
(c) 1 second, 45 m
(d) 1 second
Step-by-step explanation:
The given height as a function of height is
[tex]h= -5t^2+10t+40[/tex]
where t is the time in seconds and h is the height, in meters.
(a) At the starting time, t=0, the arrow is at the platform.
So, putting t=0, to get the height of the platform
[tex]h= -5(0)^2+10(0)+40 = 40[/tex] m
(b) For determining the length of time the arrow is in the air, first t determining the value of t for which h=0, we have
[tex]-5t^2+10t+40=0 \\\\-t^2+2t+8=0 \\\\-t^2+4t-2t+8=0 \\\\-t(t-4)-2(t-4)=0 \\\\(-t-2)(t-4)=0 \\\\[/tex]
t=-2, t=4
Mathematically, the value of t for which the arrow is in the air is the time from -2 to 4 seconds.
But the arrow was at the platform at time t=0. So, neglecting the negative time, the time for which the arrow was in the air is the time from 0 to 4 seconds, i.e 4 seconds.
(c) Mathematically, the value of t for which the arrow is in the air is the time from t=-2 to t=4 seconds. So, in the middle of time, i.e at t=1 second the arrow will reach the maximum height.
Putting t=1 to get the maximum height, we have
[tex]h=-5(1)^2+10(1)+40[/tex]
h=-5+10+40
h=45 m
Therefore, the maximum height is 45 m at t=1 second.
(d) Referinf option (c), at the time, t=1 second, the arrow has a height of exactly 45m above the ground
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