Answer:
Approximately 1.85 sec does it take the object to reach the ground
Step-by-step explanation:
As per the statement:
If an object is dropped from a height of 55 feet
⇒[tex]h_0 = 55[/tex] ft
The function is given by:
[tex]d = -16t^2+55[/tex] .....[1] gives the height of the object after t seconds.
We make the table for some values of t.
t d
0 55
1 39
2 -9
3 -89
4 -201
Plot these points on the coordinate plane.
You can see the graph as shown below.
Now, find how long does it take the object to reach the ground (d=0).
Substitute d = 0 in [1] we have;
[tex]0 = -16t^2+55[/tex]
⇒[tex]16t^2 = 55[/tex]
Divide both sides by 16 we get;
[tex]t^2 = 3.4375[/tex]
⇒
[tex]t =\pm \sqrt{3.4375}[/tex]
Since, t cannot be in negative.
⇒[tex]t = 1.85404962[/tex] second.
Therefore, Approximately 1.85 sec does it take the object to reach the ground
