When the rocket reaches the platform, then h must be equal to 0
h=0
Calculating for t using the equation:
h=-t^2+3t+10
0 = -t^2+3t+10
reversing sides:
t^2 -3t -10 = 0
By completing the squares or factoring, we can get the roots:
(t-5)(t+2) =0
t=5 or t=-2
Since time cannot be negative, therefore:
t = 5 seconds
Answer is D.
The time taken by the rocket to reach the platform is 5 seconds.
According to the question, the path model of the rocket on the computer screen is [tex]h=-t^2+3t+10[/tex] where, [tex]t[/tex] is time in seconds and [tex]h[/tex] is the height above the platform.
If the rockets reaches to the platform then, the height above the platoform for the rocket must be equals to zero.
So,
[tex]h=-t^2+3t+10\\-t^2+3t+10=0\\t^2-3t-10=0\\(t-5)(t+2)=0\\t=5;-2[/tex]
Hence, the time taken by the rocket to reach the platform is 5 seconds.
Learn more about quadratic equations here:
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