Question

A competitive cliff-diver jumps from a height of 75 feet. Find the number of feet the diver is above the ocean in 2 seconds. Evaluate for t=2
by using the formula 75−16t2
, where t
is time in seconds

Answer 1

To find the number of feet the diver is above the ocean after 2 seconds, we'll use the formula given:

\[
h(t) = 75 - 16t^2
\]

where \( h(t) \) is the height in feet and \( t \) is the time in seconds.

We need to evaluate \( h(2) \):

\[
h(2) = 75 - 16(2^2)
\]

Calculating \( 2^2 \):

\[
2^2 = 4
\]

Now substituting that back into the equation:

\[
h(2) = 75 - 16(4)
\]

Calculating \( 16(4) \):

\[
16 \times 4 = 64
\]

Now substituting that back into the equation:

\[
h(2) = 75 - 64
\]

Finally, subtracting gives:

\[
h(2) = 11
\]

Therefore, the diver is **11 feet above the ocean** after 2 seconds.