A diver runs horizontally with a speed of 0.759 m/s off a platform that is 10.0 m above the water. What is his speed just before striking the water?
To find the diver's speed just before striking the water, we can use the principle of conservation of energy. The total mechanical energy of the diver remains constant throughout the jump.
The initial mechanical energy of the diver is equal to the sum of his potential energy (mgh) and his kinetic energy (0.5mv^2) before jumping. At this point, his velocity (v) is 0.759 m/s.
When the diver is just about to strike the water, his potential energy is zero since he is at the water level. Therefore, all his initial potential energy is converted into kinetic energy.
Setting the initial mechanical energy equal to the final mechanical energy, we have:
mgh + 0.5mv^2 = 0.5mv_f^2
Where:
m = mass of the diver (which is not given here)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the platform above the water (10.0 m)
v = initial velocity of the diver (0.759 m/s)
v_f = final velocity of the diver (what we need to find)
We can rearrange the equation to solve for v_f:
v_f^2 = 2gh + v^2
Substituting the given values:
v_f^2 = 2 * 9.8 * 10 + 0.759^2
v_f^2 = 196 + 0.577
v_f^2 = 196.577
Taking the square root of both sides, we find:
v_f ≈ √(196.577)
v_f ≈ 14.0 m/s
Therefore, the diver's speed just before striking the water is approximately 14.0 m/s.