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Here, we can use the vertical problem to find the time that the

projectile is in the air. We can put this time into the horizontal problem

to find the velocity component. We can get the angle from this

component.

From the vertical problem we can get the max height...

as well as the time that the projectile is in the air.

Problem 1Friday, May 18, 2012

9:10 AM

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Putting this into the x-problem, we can solve for the angle.

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This problem really has two parts. We need to find the time it takes to

get to the cross bar, and then we need to use this time to find the

vertical position at that time.

In the x-problem

The y-position at this time can be found as


9:10 AM

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To find the clearance with respect to the crossbar, we have to subtract

the height of the crossbar.

To find whether the ball is moving up or down while crossing the

crossbar, we are more concerned with velocity than position.

Because this velocity is negative, that means that the ball is falling as itis going over the crossbar.

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This is all about trying to find where the water hits the pool. I would

guess that the walkway would have to be more than a meter wide.

The vertical problem will give the fall time for the water. Once we know

the fall time, we can find the x position of splashdown.

In the horizontal problem


9:10 AM

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This should be more than enough room to walk behind.

To make a scale model, both h and dneed to scale. We can

algebraically solve for the velocity needed.

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There is a progression to this problem.

She kicks the ball,1)

The ball hits the water at time tsplash2)

She hears the splash at some later time t.3)

We can also solve the problem in parts. Let's first find the time is takes

to hit the water. Using KE (3)

Knowing this time that the ball was in the air, we can find d. Using KE

(2).

Problem 4Tuesday, May 15, 2012

11:22 PM

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Now that we know dand h, we can find the distance that the sound

needs to travel r.

Sound only moves at a finite velocity. The time it takes to travel this

distance r is ts.

The total time needed for the player to hearthe splash is then...

Now we just solve this equation for velocity.

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This is a relative motion problem in 1D.

Velocity of motorist wrt ground

Velocity of police wrt ground

The relative velocity is the difference of these velocities.

This makes sense. I am in the police car going west and it appears that

the motorist is coming towards meit has an apparent velocity to the

east.

Problem 5Friday, August 24, 2012

2:50 PM

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The police car appears to be coming at me from behind.

Now we are talking about positions rather than velocities. We know our

separation and our relative velocity.

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This is actually the easy part of the problem. Once the ball is shot, it is

in freefall. That means there is only one acceleration in the problem

(the acceleration from gravity g) and it is always directed down.

This is the more complicated projectile motion problem. From the x-

problem, we can find the time that the ball is in the air.

We plug this time into the vertical problem.


9:10 AM

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Solving this for velocity solves the problem.

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This part is a simple kinematics problem.

This is also a pretty straight-forward kinematics problem.

Problem 7Frida y, May 18, 2012

9:10 AM

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When it reaches the edge of the cliff, the car has an initial x velocity and

an initial y velocity.

The x-component of velocity isn't going to change. The y-component of

velocity is. We can use KE (3) to find the time it takes for the car to get

to the water (we will need this time later).

Solving this quadratic equation gives two times

Knowing this time gives us an easy way to find the final components of

velocity.

The magnitude of velocity entering the water is then

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The angle is found in the normal way

Knowing the time on the ledge and the time in the air, we get the totaltime that the car is in motion.

Because the x velocity is constant and we know the time that the car is

in the air, it is simply

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In general, for cases of constant acceleration, we have...

Once we know the acceleration, we can find the components of force.

Putting them together...

Problem 8Monday, August 27, 2012

2:20 PM

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We have 2 free-body diagrams in this problem.

We have 2 equations and 2 unknowns so we can solve this uniquely.


9:10 AM

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Obviously, the first string to break in this problem will be T1. If we set T1

to the breaking tension, we can solve for the associated acceleration.

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We did this type of problem in class. Let us assume that m1 has an

acceleration that is upward. We can then solve for the required force

when setting this acceleration equal to zero.

Plugging the blue equation into the green one...


9:10 AM

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By setting a equal to zero, the minimum that Fxcan be without m1

accelerating upwards is...

For this, we will have to accelerate m2 to the left and the accelerations

will be swapped.

Plugging the blue equation into the green equation...

If the tension goes to zero, the minimum force required is

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This is obviously a force to the left.

HW 2 Solns

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