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 point)

1 answer

To find the height of the diver above the ocean after 2 seconds using the formula \( h(t) = 75 - 16t^2 \), we need to substitute \( t = 2 \) into the equation.

Given:

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

Substituting \( t = 2 \):

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

Calculating \( 2^2 \):

\[ 2^2 = 4 \]

Now substituting back into the equation:

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

Calculating \( 16 \times 4 \):

\[ 16 \times 4 = 64 \]

Now substituting this value back:

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

Finally, performing the subtraction:

\[ h(2) = 11 \]

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