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 the time in seconds.

1 answer

To find the number of feet the diver is above the ocean at \( t = 2 \) seconds, we can plug \( t = 2 \) into the formula:

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

Substituting \( t = 2 \):

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

Calculating \( 2^2 \):

\[ 2^2 = 4 \]

Now substitute this back into the equation:

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

Calculating \( 16 \times 4 \):

\[ 16 \times 4 = 64 \]

Now substitute this value back into the equation:

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

Finally, perform the subtraction:

\[ h(2) = 11 \]

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