Answer:
The height h of an object after t seconds is
h=-16t^2+48t+210h=−16t
2
+48t+210
The height of a neighboring 50-foot tall building is modeled by the equation h=50.
The time (t) when the object will be at the same height as the building is found to be t = –2 and t = 5.
To find:
The statement which describes the validity of these solutions.
Solution:
We have,
h=-16t^2+48t+210h=−16t
2
+48t+210
Here, t is the time in seconds.
For t=-2,
h=-16(2)^2+48(-2)+210h=−16(2)
2
+48(−2)+210
h=-64-96+210h=−64−96+210
h=50h=50
For t=5,
h=-16(5)^2+48(5)+210h=−16(5)
2
+48(5)+210
h=-400+240+210h=−400+240+210
h=50h=50
So, the value of h is 50 at t=-2 and t=5.
We know that time is always positive so it cannot be negative value. It means t=-2 is not possible.
The solution t = 5 is the only valid solution to this system since time cannot be negative.
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