A 69 kg diver jumps off a 10.4 m tower.
(a) Find the diver's velocity when he hits the water.
m/s
(b) The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.
N
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To solve this problem, we can use the principles of kinematics and dynamics. Let's break it down into two parts:
(a) Finding the diver's velocity when he hits the water.
To determine the diver's velocity when he hits the water, we can use the following kinematic equation:
v² = u² + 2as
Where:
v is the final velocity (unknown)
u is the initial velocity (0 m/s since the diver jumps vertically downward)
a is the acceleration due to gravity (-9.8 m/s², assuming we neglect air resistance)
s is the displacement (10.4 m, the height of the tower)
Rearranging the equation to solve for v:
v² = 0² + 2(-9.8)(10.4)
v² = -2(9.8)(10.4)
v² = -203.84
v ≈ -14.28 m/s
Since the problem specifies that the velocity is downward, we use a negative sign. So the diver's velocity when he hits the water is approximately -14.28 m/s (downward).
(b) Finding the net force exerted by the water.
To determine the net force exerted by the water, we can use Newton's second law of motion:
F = ma
Where:
F is the net force (unknown)
m is the mass of the diver (69 kg, given)
a is the acceleration (unknown)
The diver comes to a stop, which means his final velocity is 0 m/s. We need to find the acceleration in order to calculate the net force. Using the kinematic equation:
v² = u² + 2as
As the final velocity (v) is 0 m/s, and we know the initial velocity (u) is -14.28 m/s, and the displacement (s) is -2.0 m (since the diver stops 2.0 m below the surface), we can substitute these values into the equation:
0² = (-14.28)² + 2a(-2.0)
Simplifying:
0 = 203.84 - 4a
Rearranging the equation to solve for a:
4a = 203.84
a = 203.84/4
a = 50.96 m/s²
Now that we have the acceleration, we can find the net force by substituting into Newton's second law of motion equation:
F = m * a
F = 69 kg * 50.96 m/s²
F ≈ 3,514.24 N
Therefore, the net force exerted by the water on the diver is approximately 3,514.24 N.