A 65 kg diver jumps off of a 10.0 m tower.
a. Find the diver's velocity when he hits the water.
b. The diver comes to a stop 2.0 m below the surface. Find the net force exerted by the water.
Ok, im a little confused about this problem. We knoe the mass of the diver is 65 kg and the distance is 10 m.To determine the velocity wouldn't you need the time too?
Thank you!!! ah where did time go 2009 last post
For the velocity, v, on contact with water, we can equate kinetic and potential energies while ignoring air-resistance:
h=10 m
g=9.8 m/s/s
(1/2)mv²=mgh
Solve for v.
For the force, it would be similar to the car acceleration problem, except that to find the deceleration in water, we would use
v1²-v²=2aS
S=distance (2 m)
v1=final velocity =0
v=initial velocity as found above
a=deceleration due to water (negative) including effects of buoyancy and gravity.
and finally
F=ma
What they call the "net force" should be the "average force". The net force changes with time during deceleration. At the low point of the dive, it equals the buoyancy, but it starts out much higher at the time of impact with the water.
A quicker way to get the average force, F, on the diver under water would be to equate the work done against the water to the initial potential energy before the jump. This P. E. should include the potential energy decrease while in the water. Thus:
M g *(10m + 2m) = F * (2 m)
F = (M g)*12/2 = 6 M g
test
lmao old
To determine the velocity of the diver when he hits the water, we can use the principles of motion. Specifically, we can use the equation of motion for freefall:
v² = u² + 2as
where:
v is the final velocity (what we want to find)
u is the initial velocity (which is 0 in this case, since the diver starts from rest)
a is the acceleration due to gravity (approximately 9.8 m/s²)
s is the vertical distance traveled by the diver (which is 10 m in this case)
Let's plug in the values into the equation:
v² = 0² + 2 * 9.8 * 10
v² = 196
v ≈ √196
v ≈ 14 m/s
Therefore, the velocity of the diver when he hits the water is approximately 14 m/s.
Now, let's move on to part b.
To find the net force exerted by the water, we can use Newton's second law of motion:
Fnet = ma
where:
Fnet is the net force (what we want to find)
m is the mass of the diver (given as 65 kg)
a is the acceleration (which we can find using the kinematic equation of motion)
Since the diver comes to a stop 2.0 m below the surface, we can consider this as the final position (s = 2 m). We can use a similar equation of motion as before:
v² = u² + 2as
However, this time we want to find the acceleration (a), so let's rearrange the equation:
a = (v² - u²) / (2s)
Substituting the values:
a = (0² - 14²) / (2 * -2)
a = -196 / -4
a = 49 m/s²
Now, let's find the net force:
Fnet = ma
Fnet = 65 kg * 49 m/s²
Fnet ≈ 3185 N
Therefore, the net force exerted by the water on the diver is approximately 3185 Newtons.