Force and Momentum of diving board

[JLPhysics]

I wanted to post this in Homework but for some reason I can't.
Does anyone know the formulas for these problems?

An 82 kg man drops from rest on a diving board 3 m above the surface of the water and comes to rest .55 s after reaching the water. What force does the water exert on him?

A .40 kg soccer ball aproaches a football player horizontally with a velocity of 18 m/s to the north. The player strikes the ball and causes it to move in the opposite direction with a velocity of 22 m/s. What impulse was delivered to the ball bythe player?

A .50 kg object is at rest. A 3 N force to the right acts on the object during a time interval of 1.50 s. What is the velocity of this object at the end of this interval? At the end of this interval a constant force of 4 N to the left is applied for 3 s. What is the velocity at the end of the 3 s?

 

[Astronuc]

One needs to show some work and effort when requesting assistance.

Here is a reference on impulse - http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html

In the first problem, an 82 kg man falls 3 m, so one calculates the kinetic energy by conservation of energy. From KE, one determines the velocity. Then in the water the man slows to rest (V=0) in 0.55 s.

Please try the other problems.

 

[ogb p]

For the first problem, we can use the formula F=ma to calculate the force exerted by the water on the man. We know the mass of the man (82 kg) and the acceleration (g=9.8 m/s^2), so we can rearrange the formula to solve for force: F=ma=82 kg x 9.8 m/s^2 = 803.6 N. This is the force exerted by the water on the man as he comes to rest in the water.

For the second problem, we can use the impulse-momentum theorem, which states that the impulse delivered to an object is equal to the change in momentum. We know the initial momentum of the ball (p=mv=0.40 kg x 18 m/s= 7.2 kg*m/s) and the final momentum (p=mv=0.40 kg x 22 m/s= 8.8 kg*m/s). The change in momentum is the difference between the final and initial momentums (8.8 kg*m/s - 7.2 kg*m/s = 1.6 kg*m/s). Therefore, the impulse delivered to the ball by the player is 1.6 kg*m/s.

For the third problem, we can use the formula v=at to calculate the final velocity of the object after the first 1.5 seconds. We know the acceleration (a=F/m=3 N/0.50 kg= 6 m/s^2) and the time (1.5 s), so we can plug these values into the formula: v=at=6 m/s^2 x 1.5 s = 9 m/s. This is the velocity of the object after the first 1.5 seconds.

For the second part of the problem, we can use the formula v=u+at to calculate the final velocity after the additional 3 seconds, where u is the initial velocity (9 m/s) and a is the same acceleration (6 m/s^2). We also know the time (3 s), so we can plug these values into the formula: v=u+at=9 m/s + 6 m/s^2 x 3 s = 27 m/s. Therefore, the final velocity of the object after the 3 seconds of the additional 4 N force to the left is 27 m/s.