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
⇒\(h_0 = 55\) ft
The function is given by:
\(d = -16t^2+55\) .....[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;
\(0 = -16t^2+55\)
⇒\(16t^2 = 55\)
Divide both sides by 16 we get;
\(t^2 = 3.4375\)
⇒
\(t =\pm \sqrt{3.4375}\)
Since, t cannot be in negative.
⇒\(t = 1.85404962\) second.
Therefore, Approximately 1.85 sec does it take the object to reach the ground
