A competitive cliff driver 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 75-16t 2 where t is time in seconds
To find the number of feet the cliff diver is above the ocean after 2 seconds, you can use the given formula with t = 2 seconds. The formula seems to be slightly incorrectly written or missing symbols, but it appears to be a simplified version of the formula for the position of an object under free fall due to gravity (neglecting air resistance):
Distance above the ocean = Initial height - (16 * t^2)
Here, 16 represents half of the acceleration due to gravity in feet per second squared (32 ft/s^2), modified for the calculation (since the original formula is s = ut + 1/2 at^2, where u = initial velocity, a = acceleration due to gravity, and t = time). Assuming the initial velocity of the diver (u) when jumping is 0 (just dropping without an initial push), the formula simplifies to:
Distance above the ocean = Initial height - 1/2 * g * t^2
Where:
- g is the acceleration due to gravity (approximately 32 ft/s^2)
- t is the time in seconds
Let's calculate the diver's height above the ocean after 2 seconds:
Distance above the ocean at t = 2 seconds = 75 - 16 * (2^2)
= 75 - 16 * 4
= 75 - 64
= 11 feet
So, after 2 seconds, the diver is 11 feet above the ocean.