Physics, 29.12.2020 03:30 poppy1173

21. A 50.0 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resistance of 1500 N to the diver's fall. If the diver comes to rest 5.0 m below the water's surface, what is the total distance between the diving board and the diver's stopping point underwater?

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Answer from: zdejesus4994

In order to solve the steps of the diving board the average net force is given as 1500N an is towards the divers fall.

The newton's second law that tells us that the sum of forces on a body is equal to zero. the acceleration will be positive, and the resistance force as negative since it opposes the movement. ∑F = m*a.

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Answer from: aaronolivera200161

d = 15.28 [m]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to zero.

Let's take the downward movement as positive i.e. the acceleration will be positive, and the resistance force as negative since it opposes the movement.

∑F = m*a

where:

∑F = sum of forces [N]

m = mass = 50 [kg]

a = acceleration [m/s²]

g = gravity acceleration = 9.81 [m/s²]

The forces acting on the diver are the weight down and the resistance force up.

$F_{w}-F_{r}=m*a$

where:

Fw = weight = m*g = mass by the gravity acceleration

Fr = resistance force = 1500 [N]

$(50*9.81)-1500=50*a\\-1009.5=50*a\\a=-20.19[m/s^{2} ]$

The negative sign indicates that the acceleration is upward, that is, it opposes the fall of the diver in the water and for this reason it stops when it reaches 5 meters deep.

The key to solving this problem is to understand that when the diver hits the water, he does it with an initial velocity. In this way with the following formula of kinematics we can find the initial velocity or velocity of entry on the surface of the water.

$v_{f}^{2} =v_{o}^{2} -2*a*x$

where:

Vo = initial velocity [m/s]

Vf = final velocity = 0 (The diver comes to rest = no movement)

a = acceleration = 20.19 [m/s²]

x = distance from the diver's stopping point to the water surface = 5 [m].

$0 = v_{o}^{2} -2*20.19*5\\v_{o}=\sqrt{(2*20.19*5)} \\v_{0}=14.2[m/s]$

Now we use the same expression of kinematics again. But with the difference that the final velocity is the velocity of entry into the water and the initial velocity is zero since the diver is launched without initial velocity. Acceleration is equal to gravitational acceleration.

$v_{f}^{2} =v_{o}^{2}+2*g*x_{1}\\ 14.2^{2}=0+(2*9.81*x_{1})\\201.64=19.62*x_{1}\\x_{1} = 10.28[m]$

Then the total fall distance will be 5 meters in the water plus the calculated distance in the air.

$d=5+10.28\\d = 15.27[m]$

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21. A 50.0 kg diver steps off a diving board and drops straight down into the water. The water provi...

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