Question

An 80 kg man standing erect steps off a 3.0 m high diving platform and begins to fall from rest. The man again comes to a rest 2.0 s after reaching the water. What average force did the water exert on him?

Answer 1

The average force the water exerted on the 80 kg man to bring him to rest in 2 seconds after stepping off a 3.0 m high diving platform is 306.8 N in the upward direction.

Step-by-step explanation:

The question is about an 80 kg man who steps off a 3.0 m high diving platform and comes to rest after entering the water. To find the average force the water exerted on him, we first need to determine his velocity just before he enters the water and then use the impulse-momentum theorem to find the force during the 2-second deceleration in the water.

To find the velocity before entering the water, we use the equation v = √(2gh), where g is the acceleration due to gravity (9.8 m/s2) and h is the height of the diving platform. Plugging in the values, we get v = √(2 * 9.8 * 3.0) ≈ 7.67 m/s.

Next, the change in momentum (Δp) is equal to the final momentum minus the initial momentum. Since he comes to rest, the final velocity (and hence momentum) is 0, and the initial momentum is mass * velocity or 80 kg * 7.67 m/s. So, Δp = 0 - (80 * 7.67) = -613.6 kg*m/s.

Because impulse equals the change in momentum, and impulse is also equal to average force (F) times the time (t) during which the force is applied, we have F * t = Δp. Rearranging for F, we get F = Δp / t = -613.6 kg*m/s / 2 s = -306.8 N. The negative sign indicates that the force is in the opposite direction of the initial motion, which means it's upwards.

Answer 2

To find the average force exerted by the water on an 80 kg man who falls from a 3.0 m high platform and takes 2.0 s to stop, calculate his impact velocity and the deceleration caused by the water, then apply Newton's second law.

Step-by-step explanation:

The question asks for the average force exerted by the water on an 80 kg man who comes to rest after falling into the water from a 3.0 m high diving platform and taking 2.0 s to come to a stop.

First, we calculate the velocity of the man just before he hits the water using the kinematic equation v2 = u2 + 2as, where u is the initial velocity (0 m/s), a is the acceleration due to gravity (9.8 m/s2), and s is the height (3.0 m). With this we find the velocity just before impact. Next, using the final velocity (0 m/s), this impact velocity, and the time taken to stop (2.0 s), we can use the kinematic equation v = u + at to find the deceleration caused by the water.

Finally, applying Newton's second law F = ma, we can calculate the average force the water exerted on the man during deceleration, taking the deceleration from the previous step and the man's mass (80 kg).