an athlete drops from rest from a platform 10m above the surface of a 5m deep pool. Assume that the athlete enters the water vertically and moves through the water with the constant acceleration, what is the minimum average force with water must exert on a 62 kg diver to prevent her from hitting the bottom of the pool? Express answer in newtons and also in multiples of the divers weight

You have to add the fores in the air and the water

Air: a=9.8 m/s^2 due to gravity
Force=ma, F=62(9.8)=607.6

Water:
First find the velocity in which the diver enters the water,
V^2=V0^2 + 2ay
V^2= 0 + 2(9.8)(10)
V= sqr(20*9.8)= 14m/s

Now use velocity to solve for acceleration through the water
V^2=V0^2 + 2ay
0=14^2 + 2a(5)
-14^2= 10a
a=19.6 m/s^2

Now find force through water
F=ma
F=62(19.6)=1215.2

Now add force through water and force through air
1215.2+607.6=1822.8 N

Well, let's dive into this question!

First, let's calculate the time it takes for the diver to reach the surface of the water when falling from the platform. We can use the equation:

s = ut + (1/2)at^2

where:
s = distance (10 m in this case)
u = initial velocity (0 m/s as the diver starts from rest)
a = acceleration (g = 9.8 m/s^2, constant acceleration due to gravity)
t = time (what we want to find)

Plugging in the values, we get:

10 = 0*t + (1/2)*9.8*t^2

Simplifying further, we get:

4.9t^2 = 10

Dividing both sides by 4.9, we find:

t^2 = 10/4.9

Taking the square root, we find:

t ≈ 1.43 seconds

Now that we know the time it takes for the diver to hit the water, we can calculate the average force exerted by the water using Newton's second law, which states:

F = ma

where:
F = force (what we want to find)
m = mass of the diver (62 kg)
a = acceleration (which we can calculate)

To calculate the acceleration of the diver in the water, we can use the equation:

s = ut + (1/2)at^2

but this time, the distance traveled is the depth of the pool (5 m), the acceleration is what we want to find, and the time is the same as before (1.43 seconds).

Plugging in the values, we get:

5 = 0*1.43 + (1/2)*a*(1.43)^2

Simplifying further:

a*(1.43)^2 = 10

Dividing both sides by (1.43)^2, we find:

a ≈ 10/(1.43)^2

Calculating this value, we find:

a ≈ 4.97 m/s^2

Now, we can find the average force using Newton's second law:

F = ma
F = (62 kg)*(4.97 m/s^2)

Calculating this, we find:

F ≈ 307.14 N

So, the minimum average force that the water must exert on the diver to prevent her from hitting the bottom of the pool is approximately 307.14 Newtons.

Now, let's express this answer in multiples of the diver's weight. The diver's weight can be calculated using the formula:

Weight = mass * gravity

Substituting the given values, we find:

Weight = 62 kg * 9.8 m/s^2

Calculating this, we find:

Weight ≈ 607.6 N

To express the minimum average force in multiples of the diver's weight, we divide the force by the weight:

307.14 N ÷ 607.6 N ≈ 0.51

So, the answer is approximately 0.51 times the diver's weight.

To find the minimum average force exerted by the water on the diver to prevent her from hitting the bottom of the pool, we need to consider the three phases of the diver's motion: the free fall (in air), the entry into the water, and the motion through the water.

Let's break down the problem step by step.

Step 1: Calculate the time taken for the diver to reach the water surface.
Using the equation of motion:
s = ut + (1/2)at^2

Here,
s = 10 m (distance from the platform to the water surface)
u = 0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity)
t = ?

Plugging in the values, we get:
10 = 0 * t + (1/2) * 9.8 * t^2
10 = 4.9 * t^2

Solving the equation, we find:
t^2 = 10/4.9
t ≈ 1.43 seconds

So, it takes approximately 1.43 seconds for the diver to reach the water surface.

Step 2: Calculate the velocity of the diver just before entering the water using the equation:
v = u + at

Here,
u = 0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity)
t = 1.43 seconds (time taken to reach the water surface)
v = ?

Plugging in the values, we get:
v = 0 + 9.8 * 1.43
v ≈ 14 m/s

The velocity of the diver just before entering the water is approximately 14 m/s.

Step 3: Determine the deceleration experienced by the diver in the water.
The deceleration depends on the depth of the water the diver enters. In this case, the water is 5 meters deep.

We can use the third equation of motion:
v^2 = u^2 + 2as

Here,
v = 0 m/s (final velocity)
u = 14 m/s (initial velocity)
a = ? (deceleration)
s = 5 m (depth of the water)

Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
a = (0 - 14^2) / (2 * 5)
a ≈ -19.6 m/s^2

The deceleration experienced by the diver in water is approximately -19.6 m/s^2.

Step 4: Calculate the force exerted by the water on the diver to prevent her from hitting the bottom.
Using Newton's second law:
F = ma

Here,
m = 62 kg (mass of the diver)
a = -19.6 m/s^2 (deceleration experienced by the diver in water)
F = ?

Plugging in the values, we get:
F = 62 * (-19.6)
F ≈ -1212.8 N

The force exerted by the water on the diver is approximately -1212.8 N.

To express the answer in multiples of the diver's weight:
Weight of the diver = mg
Weight = 62 kg * 9.8 m/s^2 ≈ 608 N
Force = -1212.8 N

Dividing the force by the weight:
-1212.8 N / 608 N ≈ -2

Therefore, the minimum average force with which the water must exert on the 62 kg diver to prevent her from hitting the bottom of the pool is approximately -1212.8 N or -2 times her weight. (Note: The negative sign indicates that the force is exerted upward to counter the downward motion.)

m•g•h =m•v²/2,

v² =2•g•h,
a=v²/2•s=2•g•h/2•s= g•h/s=9.8•10/5=2•g=19.6 m/s².
F=m•a= m•2g= 2(m•g)=2•62•19.6 = 1215.2 N.