A high diver of mass 69.4 kg jumps off a board 9.6 m above the water.
The acceleration of gravity is 9.8 m/s2.If his downward motion is stopped 2.21 s after he enters the water, what average upward force did the water exert on him?
Answer in units of N
To find the average upward force exerted by the water on the high diver, we need to determine the change in momentum of the diver during his descent into the water.
First, let's calculate the diver's velocity just before entering the water. We can use the kinematic equation:
v = u + at
where v is the final velocity (0 m/s), u is the initial velocity, a is the acceleration due to gravity (-9.8 m/s²), and t is the time taken to enter the water (2.21 s). Rearranging the equation, we get:
u = v - at
u = 0 - (-9.8 * 2.21)
u = 0 + 21.658
u = 21.658 m/s
Now, we can calculate the diver's initial kinetic energy just before entering the water using the formula:
KE = 0.5 * m * u²
where KE is the kinetic energy, m is the mass of the diver (69.4 kg), and u is the initial velocity (21.658 m/s). Substituting the values, we have:
KE = 0.5 * 69.4 * (21.658)²
KE = 0.5 * 69.4 * 469.944964
KE = 16229.71 J
Since the diver comes to rest in the water, all of his initial kinetic energy is transferred to the water as work done against the upward force. So, the work done is equal to the initial kinetic energy.
Now, we can calculate the average upward force by dividing the work done by the distance travelled by the diver vertically. The distance is equal to the height of the board, which is 9.6 m.
Work done = force * distance
Average upward force = Work done / distance
Average upward force = 16229.71 J / 9.6 m
Average upward force ≈ 1694.96 N
Therefore, the average upward force exerted by the water on the high diver is approximately 1694.96 Newtons.