Question

If an object is dropped from a height of 55 feet, the function d=-16t^2+55 gives the height of the object after t seconds. Graph this function. Approximately how long does it take the object to reach the ground (d=0)?

Answer 1

To find the time to reach ground solve -16t^2 +55 = 0

t^2 = 55/16
t = sqrt55 / 4 = 1.85 seconds

Answer 2

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