Question

A 50.9 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resistance of 1492 N to the diver’s fall. If the diver comes to rest 6 m below the water’s surface, what is the total distance between the diving board and the diver’s stopping point underwater? The acceleration due to gravity is 9.81 m/s 2 .

Answer 1

Answer:

T = 23.92 m

Explanation:

given,

mass of the diver = 50.9 Kg

Resistant force from the water = f = 1492 N

diver come top rest under water at a distance = 6 m

acceleration due to gravity = 9.81 m/s²

final velocity = v  = 0 m/s

initial velocity = u = ?

total distance = ?

Now acceleration of body under water

f = m a

$a = \dfrac{1492}{50.9}$

$a = 29.31\ m/s^2$

using equation of motion

v² = u² + 2 a s

0 = u² - 2 x 29.31 x 6

$u = \sqrt{2\times 29.31\times 6}$

u = 18.75 m/s

now.

calculating distance of the diver in air

v² = u² + 2 a s

0 = 18.75² - 2 x 9.81 x s

s = 17.92 m

total distance

T = 17.92 + 6

T = 23.92 m

the total distance between the diving board and the diver’s stopping point underwater T = 23.92 m