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

1 answer

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.