A 751 N diver drops from a board 9.10 m
above the water’s surface.
The acceleration of gravity is 9.81 m/s2 .
a) Find the diver’s speed 3.00 m above the
water’s surface.
Answer in units of m/s.
KE final= PE initial
1/2 mv^2=mgh
solve for v.
To find the diver's speed 3.00 m above the water's surface, we can use the principle of conservation of energy. The potential energy of the diver at a given height is converted into kinetic energy as the diver falls.
First, let's calculate the potential energy of the diver at a height of 9.10 m above the water's surface. The potential energy (PE) is given by the formula:
PE = m * g * h
Where:
- m is the mass of the diver (not given in the question)
- g is the acceleration due to gravity, 9.81 m/s^2
- h is the height, 9.10 m
Since the mass of the diver is not given, we can ignore it for now since it's canceled out in the final calculation of the speed.
Now, let's calculate the potential energy of the diver at a height of 9.10 m:
PE = 751 N * 9.81 m/s^2 * 9.10 m
PE = 66062.41 N·m or J (Joules)
According to the principle of conservation of energy, the potential energy at a certain height is equal to the kinetic energy at a lower height. Therefore, the kinetic energy of the diver at a height of 3.00 m above the water's surface will also be equal to 66062.41 J.
The kinetic energy (KE) is given by the formula:
KE = 1/2 * m * v^2
Where:
- m is the mass of the diver (again, canceled out in the final calculation)
- v is the velocity or speed of the diver
Now we can solve for the velocity by rearranging the formula:
v = sqrt(2 * KE / m)
Since the mass of the diver is canceled out, we can directly calculate the velocity:
v = sqrt(2 * 66062.41 J)
v ≈ 288.9 m/s
Therefore, the diver's speed 3.00 m above the water's surface is approximately 288.9 m/s.