Question

A cliff diver jumps off of a 24 foot cliff into the ocean.

Five feet away from the edge of the cliff he reaches his
maximum height of 25 feet, and splashes into the ocean.
How long is the diver in the air according to this problem?
Is this reasonable?

Answer 1

The diver is in the air for approximately 2.55 seconds.

Step-by-step explanation:

The time the diver is in the air can be calculated using the vertical motion equations. The diver reaches his maximum height when his vertical velocity becomes zero. We can set up the equation:

• v = u + at

Where v is the final velocity at the highest point, u is the initial velocity, a is the acceleration due to gravity, and t is the time the diver is in the air. In this case, the final velocity is 0 m/s, the initial velocity is the velocity at which the diver reaches the maximum height, which is 25 feet, and the acceleration due to gravity is -9.8 m/s² (taking into account that the acceleration due to gravity acts downward).

Plugging in the values we have:
0 = 25 + (-9.8)t

Solving for t:

t = 25/9.8 ≈ 2.55 seconds

This means that the diver is in the air for approximately 2.55 seconds.