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 number of feet the diver is above the ocean after 2 seconds using the formula \( h(t) = 75 - 16t^2 \), we will substitute \( t = 2 \) into the formula.

  1. Start with the equation: \[ h(t) = 75 - 16t^2 \]

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

  3. Calculate \( 2^2 \): \[ 2^2 = 4 \]

  4. Now multiply by 16: \[ 16 \cdot 4 = 64 \]

  5. Substitute this back into the equation: \[ h(2) = 75 - 64 \]

  6. Finally, calculate: \[ h(2) = 11 \]

Thus, the diver is 11 feet above the ocean at \( t = 2 \) seconds.