Physics help

[hotcommodity]

Well, I have all summer to prepare for calculus based physics in the fall. I started last night and I've already come upon a problem. I'll just use this thread to ask questions instead of making 100 of them throughout the summer. Any help is appreciated.

 

The problem is: A man throws a ball horizontally such that the ball leaves 1.5m above the ground. At what velocity must the ball leave his hand so that it reaches a point 20m away without touching the ground before it gets to that point?

 

I got that the initial velocity in the y-direction is 0, which allowed me to find t. But that still leaves acceleration in the x direction and initial velocity in the x direction to be accounted for. I'm not sure where to go from here.

[NeonBlack]

After the ball has been thrown, how many forces are in the x-direction?

[hotcommodity]

After its been thrown, none i think....

[NeonBlack]

Right. Exactly zero forces.

[hotcommodity]

I'm not sure what that infers. No acceleration in the x direction?

[Klaynos]

I'm not sure what that infers. No acceleration in the x direction?

 

F=ma=0 either a=0 or m=0

[hotcommodity]

I got that t=2s, and x=20m, so v-initial would be 10 m/s. The books answer is 36.1 m/s. I'm not sure where I went wrong.

[Klaynos]

I think you have your time wrong.

 

How did you work it out?

[hotcommodity]

Ah ok, I see where I went wrong. I was using the x distance for figuing out the time instead of the y distance. Thanks for everyones help.

[Klaynos]

Ah ok, I see where I went wrong. I was using the x distance for figuing out the time instead of the y distance. Thanks for everyones help.

 

np, very common mistake that, get a time of about 0.5s now?

[hotcommodity]

np, very common mistake that, get a time of about 0.5s now?

 

Yep, I did. Sometimes my book is wrong so its hard to know if its me or the book, lol. But thanks again for your help

[NeonBlack]

Ah ok, I see where I went wrong. I was using the x distance for figuing out the time instead of the y distance. Thanks for everyones help.

 

I suggest always drawing a picture a free-body diagram with your problems, even if it seems trivial. If you had done this, you probably would have immediately realized that there is no x-accelaration and also avoided this mistake.

[hotcommodity]

I suggest always drawing a picture a free-body diagram with your problems, even if it seems trivial. If you had done this, you probably would have immediately realized that there is no x-accelaration and also avoided this mistake.

 

I did, I just have very little experience with physics.

 

Another problem: A cannon shoots a ball at 45 degrees from the horizontal. The ball travels 80 miles and hits with the same vertical speed that it had when it left the cannon. Find the time interval and the initial velocity.

 

I have that [math]a_x=0[/math], [math]x=1.3E5m[/math], and the initial speed of the ball in the y-direction is equal to the final speed in the y-direction. Every time I plug the numbers into the kinematics eq.'s I still have two unknowns. I'm not sure where to go from here.

[swansont]

I did, I just have very little experience with physics.

 

Another problem: A cannon shoots a ball at 45 degrees from the horizontal. The ball travels 80 miles and hits with the same vertical speed that it had when it left the cannon. Find the time interval and the initial velocity.

 

I have that [math]a_x=0[/math], [math]x=1.3E5m[/math], and the initial speed of the ball in the y-direction is equal to the final speed in the y-direction. Every time I plug the numbers into the kinematics eq.'s I still have two unknowns. I'm not sure where to go from here.

 

Given the 45 degrees, can you relate the initial x and y speeds to each other?

 

Also, FYI, the "same final speed as initial" is also code for "no air resistance"

[hotcommodity]

Given the 45 degrees, can you relate the initial x and y speeds to each other?

 

Also, FYI, the "same final speed as initial" is also code for "no air resistance"

 

Ok, I got it. I see that the initial speeds are the same. Thank you

[hotcommodity]

Well, I've come upon a few more problems, any help is appreciated.

 

I wanted to ask how something can move at a constant velocity, and have no net force acting upon it. If the object is still, it makes sense, but if someone is draging a sled for instance at a constant velocity, how do the forces balance out to give zero net force? I assume kinetic friction plays a role in that, but what if the space shuttle is moving through space at a constant velocity, what cancels out the thrust such that it would not accelerate?

 

I'll post the other problems later, thanks for any help...

[D H]

but what if the space shuttle is moving through space at a constant velocity, what cancels out the thrust such that it would not accelerate?

 

The shuttle will accelerate if it is firing its thrusters. For example, the shuttle main engines stops firing when the shuttle is about 150 miles per hour short of orbital velocity. This is the main engine cutoff (MECO). The shuttle separates from the external tank after MECO. The shuttle and external tank rise to apogee with no engines firing. At apogee, the shuttle fires its orbital maneuvering system jets to give it that extra 150 mph and then stops firing again: it is in orbit and doesn't need to fire its thrusters any more. The external tank doesn't have any thrusters; it just falls into the Indian Ocean.

[hotcommodity]

Ok, here are the other questions I'm stuck on...

 

1. An airplane is flying with a velocity of 240 m/s at an angle of 30 degrees with the horizonal. When the altitude of the plane is 2.4 km, a flare is released from the plane. The flare hits the target on the ground. What is the angle theta?

 

And the book has this drawing:

 

 

So I used the equation [math]y= v_0y*t + (1/2)*a_y*t^2[/math] to get [math]-2400 = (240*sin(30))*t - 4.9*t^2[/math]. Sovling for t I get 37.5 degrees. I multiply t times (240*cos(30)) to get the x-distance, which I think will give me the length of the the hypotenuse. knowing the length of these two sides, I find the angle to be 17 degrees, which is wrong. The answer is 42 degrees. I think I'm using the wrong angle of 30 degrees, but I dont know.

 

2. A police car is traveling at a velocity of 18.0 m/s due north, when a car speeds by at a constant velocity of 42.0 m/s dur north. After a reaction time of .800 s, the policeman begins to pursue the speeder at an acceleration of 5 .00 m/s^2. Including the reaction time, how long does it take for the policeman to catch up with the speeder?

 

I used the eq. [math]y= v_0y*t + (1/2)*a_y*t^2[/math] for both the policeman and the speeder, and set them equal to eachother, because I figure that their distances have to equal eachother when the cop catches up to the speeder. So I have [math]18*t + 2.5*t^2 = 42*t[/math], and I solved for t to get 9.6 s. The 9.6 plus the reaction time would give me 10.4 s. The answer is supposed to be 11.1 s.

 

3. A duck has a mass of 2.5 kg. As the duck paddles, a force of .10 N acts on it in a direction due east. In addition, the current of the water exerts a force of .20 N in a direction of 52 degrees south of east. When these forces begin to act, the velocity of the duck is .11 m/s in a direction due east. Fine the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 3.0 s while the forces are acting.

 

Here I found the x and y component forces first. In the x-direction, .10 N from the paddling, and .20*cos(52) from the wind, added together it gives .223 N. In the y-direction, theres only a force of -.20*sin(52), which gives -.158 N. Adding the two forces squared and taking the root to get the net force, I have .273 N. To get the acceleration, I divide that by the ducks mass to get .109 m/s^2. Then I have [math] x= v_0x*t + (1/2)*a_x*t^2= .11*3 + .5*.109*9= .82 m[/math]. The book has .78 m.

 

Lastly...

 

4. A person has a chioce of either pushing or pulling a sled at a constant velocity, as the drawing illustrates. Friction is present. If the angle theta is the same in both cases, does it require less force to push or pull?

 

The book has this picture:

 

 

I would think that it would take less force to pull the sled, because the y-component from the pushing is just pointing into the ground, but I'm not sure.

 

@ D H

 

Thanks for the reply. So are you saying that the shuttle wouldn't be able to move at a constant velocity? I'm still wondering in the case of the sled being pulled at a constant velocity if it is the kinetic friction that causes there to be no net force...Does no net force always mean that the forces acting on an object are balanced out?

[swansont]

2. A police car is traveling at a velocity of 18.0 m/s due north, when a car speeds by at a constant velocity of 42.0 m/s dur north. After a reaction time of .800 s, the policeman begins to pursue the speeder at an acceleration of 5 .00 m/s^2. Including the reaction time, how long does it take for the policeman to catch up with the speeder?

 

I used the eq. [math]y= v_0y*t + (1/2)*a_y*t^2[/math] for both the policeman and the speeder, and set them equal to eachother, because I figure that their distances have to equal eachother when the cop catches up to the speeder. So I have [math]18*t + 2.5*t^2 = 42*t[/math], and I solved for t to get 9.6 s. The 9.6 plus the reaction time would give me 10.4 s. The answer is supposed to be 11.1 s.

 

You've ignored the distance travelled by the two during the "reaction time," which isn't equal, so your equation (as written) isn't true — the policeman travels further during the time t, during which the cop accelerates.

[D H]

1. So I used the equation [math]y= v_0y*t + (1/2)*a_y*t^2[/math] to get [math]-2400 = (240*sin(30))*t - 4.9*t^2[/math]. Sovling for t I get 37.5 degrees.

 

You have a sign error here.

 

3. Here I found the x and y component forces first.

You should do that all the way through to the end. What you did here is invalid (you are treating the problem as one dimensional):

Then I have [math] x= v_0x*t + (1/2)*a_x*t^2= .11*3 + .5*.109*9= .82 m[/math].

[hotcommodity]

First off, I appreciate the replies above

 

I have a question about uniform circular motion. I was wondering to myself whether I could calculate the period of the Earths rotation from the mass of an object on Earth. I was using Newtons Universal Law of Gravitation for the centripetal force acting on the mass and the other basic equations involving tangential velocity, radius, and period. I kept getting strange numbers for my answer, nothing close to the period of the Earths rotation, and my reasoning is this: I cannot calculate the period of the Earths rotation by using an object of mass that exists on the Earth because it is not the rotation of the Earth that pulls us toward its center. The gravitation field is what keeps us attracted to earth, and even though the Earth rotates, the centripetal force that acts on the objects on earth is negligible. Is my reasoning correct?

[swansont]

First off, I appreciate the replies above

 

I have a question about uniform circular motion. I was wondering to myself whether I could calculate the period of the Earths rotation from the mass of an object on Earth. I was using Newtons Universal Law of Gravitation for the centripetal force acting on the mass and the other basic equations involving tangential velocity, radius, and period. I kept getting strange numbers for my answer, nothing close to the period of the Earths rotation, and my reasoning is this: I cannot calculate the period of the Earths rotation by using an object of mass that exists on the Earth because it is not the rotation of the Earth that pulls us toward its center. The gravitation field is what keeps us attracted to earth, and even though the Earth rotates, the centripetal force that acts on the objects on earth is negligible. Is my reasoning correct?

 

Right. What you'd want is the difference between the gravitational force and the normal force on a body resting on the surface. If the gravitational force were acting completely as the centripetal force, there would be no normal force, i.e. we'd be in orbit around the earth.

 

What you are probably calculating is the rotation period for that to occur.

[hotcommodity]

I see, that makes more sense, thank you

[hotcommodity]

I have another question about uniform circular motion. I was plugging in numbers to the equations that describe tangential velocity, radius, and period and found some differences between the gravitational force and other centripetal forces such as tension. For a centripetal force like tension, I found that:

 

1. As the radius increases, the tangetial velocity increases.

 

2. As the radius increases, the centripetal acceleration increases.

 

3. As the tangetial velocity increases, the centripetal acceleration increases.

 

But for a centripetal force caused by gravitation:

 

1. As the radius increases, the tangential velocity decreases.

 

2. As the radius increases, the centripetal acceleration decreases.

 

3. As the tangetial velocity increases, the centripetal acceleration increases.

 

Number 3 is the same in both cases, but 1 and 2 differ. Can this difference be attributed to the gravitational field?

[swansont]

The two cases are the force being constant and the force varying as 1/r² which is why they behave differently. Nothing magical about it being gravity; the electrostatic force from a point charge would behave the same way.