Answer: The speed of the diver's fall will be approximately 17.15 m/s.
Explanation:
We'll use the equation:
v^2 = u^2 + 2as
Where:
v is the final velocity (speed of the diver's fall)
u is the initial velocity (which is 0 m/s)
a is the acceleration due to gravity (9.8 m/s²)
s is the distance fallen (15 meters)
Substituting the given values:
v^2 = 0^2 + 2 x 9.8 x 15
v^2 = 0 + 2 x 9.8 x 15
v^2 = 2 x 9.8 x 15
v^2 = 294
Taking the square root of both sides to solve for v:
v = √294
v ≈ 17.15 m/s.
So, the speed of the diver's fall will be approximately 17.15 m/s.
