physics help

I need help on solving the following problem: A ball is thrown from a height of 1.6m, at a speed of 11.5m/s and at an angle of elevation of 45.9 degrees. How far from the person who threw it does the ball land?

Originally posted by wormer
I need help on solving the following problem: A ball is thrown from a height of 1.6m, at a speed of 11.5m/s and at an angle of elevation of 45.9 degrees. How far from the person who threw it does the ball land?

I recommend you visit www.cramster.com and sign up. You ask your questions there, and will probably have more people answer you than on a chess site 🙂

Sign up there and post your question in the physics forum, and then I'll answer it there (so I get some credit) unless someone beats me to it.

Actually, with such a basic question as that, I'd say your question will be fully answered within 10 - 30 minutes of asking it if you go to cramster.com. Even if someone answers it here - go to cramster and sign up an excellent site for 'studying' (most people just go there, post their homework questions, and then copy down the solutions, don't tell anyone there I said that)

Originally posted by wormer
I need help on solving the following problem: A ball is thrown from a height of 1.6m, at a speed of 11.5m/s and at an angle of elevation of 45.9 degrees. How far from the person who threw it does the ball land?

Whenever you're having trouble with a physics problem, try following these basic steps:

1. Draw a picture of the situation. Make sure you include all the pertinent information (the height of the ball when thrown, the shape of the path the ball will take, a vector indicating the initial speed, the angle at which is was thrown, etc...).

2. Decide which variables and equations are important. If you're stuck, try asking yourself what would change the answer. What would happen if the ball were thrown harder? What about if it were thrown from higher up? What if it were thrown on a steeper angle? A shallower angle? What if it were thrown on the moon? In outer space? By asking yourself these questions, and visualizing the answers, you'll help develop your physical intuition (the thing that lets you know when you've made a mathematical mistake, because the result is unrealistic). Once you've discovered which variables matter, review your basic equations and see which ones contain the variables you're interested in. Then ask yourself if they apply in this case. Do the equations help you determine specific points along the path of a thrown object? Or do they only describe the energy of the projectile?

3. Decide what you're looking for. In this case, you're looking for distance, so find the equation that relates distance to the important variables (your "master equation"😉. Determine if there are any variables in the equation that are currently unknown, and if there are try to apply another equation that describes the unknown variable. Keep working backwards like this until you have all the information you need to solve the master equation.

4. Crunch the numbers. If algebra isn't your forte, make sure to calculate one step at a time and write each step out. It make take you an extra minute or two, but the effort will be well worth it when you look back on your solution and see how clear it is.

5. Check your answer. First, ask yourself if the answer seems physically realistic. If you come up with a distance of 2 m (about 6 ft) or 200 m (600 ft), you've probably made a mistake - when's the last time you saw someone throw a ball that short or that far? If you come up with a reasonable answer, try plugging it back into your equations and solving for a given variable (e.g. try plugging in the distance and solving for the initial velocity). If the answers match up, chances are you're doing fine.

6. Answer the question. The easiest way to make sure you've answered the correct question is to paraphrase the original. In this case, the question is "How far from the person who threw it does the ball land?", so an appropriate answer would be something like "The ball landed x m from the person who threw it.". Remember, physics is about predicting and explaining the physical world around us, so words count. The math is simply a tool that serves this ultimate purpose.

Try this method out, and feel free to post your work in progress. Remember, if you get stuck ask for help! Just make sure you've put some thought into the problem and identified a specific stumbling block before you ask, because having someone serve you the answer on a silver platter does you absolutely no good.

Good luck!

I need help trying to figure out the angle going downward. I understand that trigonometry must be used.

Originally posted by wormer
I need help on solving the following problem: A ball is thrown from a height of 1.6m, at a speed of 11.5m/s and at an angle of elevation of 45.9 degrees. How far from the person who threw it does the ball land?

To start, you should know that the motion of the ball should be analyzed in two separate steps: vertical and horizontal. The starting vertical velocity is 11.5 m/s x sin 45.9 degrees, and the starting horizontal velocity is 11.5 m/s x cos 45.9 degrees.

Then you figure out how long the ball takes to hit the ground vertically, and that lets you know how long the ball is flying horizontally.

That's not the whole solution, but it's a strong start.

Originally posted by wormer
I need help trying to figure out the angle going downward. I understand that trigonometry must be used.

You don't need to know any downward angles for this problem. You need to calculate the time it takes to hit the ground.

Originally posted by wormer
I need help trying to figure out the angle going downward. I understand that trigonometry must be used.

let the sides of the triangle be a,b,c, and the corresponding angles are alpha, beta, gamma.

you want to break the velocity vector into its horizontal and vertical components.

to get the vertical component(ie the component of the velocity that limits the time flight, which in turn governs the horizontal distance), use "SOH"

sin(alpha)= a/c

you have c, and the angle "alpha", so solve for the leg "a"

use a similiar approach to solve for the horizooontal component of velocity

As I said earlier, head to cramster.com and you will receive the full solution. You would / should have already received the solution had you posted in cramster yesterday.

Originally posted by Ramned
As I said earlier, head to cramster.com and you will receive the full solution. You would / should have already received the solution had you posted in cramster yesterday.

I have gotten plenty of help from this site, as you can see that there are plenty of bright enough people on here to answer your question.

use the kinematic equations for constant acceleration, as a function of time

y=1/2*g*t^2 + Voy*t + yo

y=final position
g= aaceleration due to gravity
t=time
Voy=the vertical component of the initial velocity
yo=inital hieght

Voy=11.5m/s*sin(45.9deg)

now that all of the values, except one are known( the time) solve for it

use the quadratic formula, eliminating the negative solution.

once you get a value for time, multiply it by the horizontal component of the initial velocity( ie 11.5m/s*cos(45.9deg))

Vox*t = horizontal displacement

Originally posted by joe shmo
I have gotten plenty of help from this site, as you can see that there are plenty of bright enough people on here to answer your question.

Of course there are experienced people here. But they come to play chess more than solve physics problems. People go to cramster with the purpose of solving other people's problems. You'll get better help from the cramster people (i.e. myself). People also tend to give the correct answers 😉 (Your method is wrong)

Originally posted by Ramned
Of course there are experienced people here. But they come to play chess more than solve physics problems. People go to cramster with the purpose of solving other people's problems. You'll get better help from the cramster people (i.e. myself). People also tend to give the [b]correct answers 😉 (Your method is wrong)[/b]

oh...perhaps you will do the favor of pointing out my error....by the way, I also handle packages for UPS..funny

Originally posted by joe shmo
I have gotten plenty of help from this site, as you can see that there are plenty of bright enough people on here to answer your question.

use the kinematic equations for constant acceleration, as a function of time

y=1/2*g*t^2 + Voy*t + yo

y=final position
g= aaceleration due to gravity
t=time
Voy=the vertical component of the initial velocity ...[text shortened]... l component of the initial velocity( ie 11.5m/s*cos(45.9deg))

Vox*t = horizontal displacement

by this you are making the assumption that it is stopped at the height it is thrown at, and then you add the initial height?

Originally posted by wormer
by this you are making the assumption that it is stopped at the height it is thrown at, and then you add the initial height?

how do you know that my hieght of 1.6m wasnt my "yo"=0, thus leaving the final position of y=-1.6m?

Joe your solution isn't right.

1) Find the X and Y components of initial velocity.

Y component is Vsin(angle)
X Component is Vcos(angle)

2) Find the time it takes to reach its peak.

Vpeak = Vy + gT
Vpeak = Vsin(angle) + gT
0 = Vsin(angle) + -9.8T

*At the peak, the vertical velocity is 0 m/s.

Solve for T. This is the time it takes to get to peak.

3) Double the time that it takes to get to peak, because the ball falls back down.

This time is now how long the ball takes to fall back down to its initial launch height.

4) Now it still falls 1.6 meters. Figure out how long it takes for the ball to fall 1.6 meters. What are your givens?

You know the velocity at the 1.6 meter mark is equal and opposite to the given launch speed (kinematics is symmetrical!)

Gravity = -9.8 m/s^2

Use:

Y = VyT + .5gT^2
-1.6 = Vsin(angle) + .5(-9.8)(T^2) [see step 1 for what is Vsin(angle)]

You can find the time now using quadratics.

5) You now have the time it's in air.

D = VxT
D = Vcos(angle) * Time in Air

Solve from there. Again - > please check cramster.com out, much more efficient and the solutions are easier to follow. In fact some people there will give you the numerical answer.

Originally posted by Ramned
Joe your solution isn't right.

1) Find the X and Y components of initial velocity.

Y component is Vsin(angle)
X Component is Vcos(angle)

2) Find the time it takes to reach its peak.

Vpeak = Vy + gT
Vpeak = Vsin(angle) + gT
0 = Vsin(angle) + -9.8T

*At the peak, the vertical velocity is 0 m/s.

Solve for T. This is the time it ta solutions are easier to follow. In fact some people there will give you the numerical answer.

The entire thing until between step 4 and 5 is un-necessary

If you pay attention, step 5 is the exact equation I derived.

let me ask you this, why do you feel it is necessary to find the max height. It is irrelevant to this problem.?