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Projectile motion of ball at wall problem

  1. Oct 15, 2006 #1
    Okay I know this should be easy but for some reason I just cant get it.

    "You throw a ball toward a wall with a speed of 25.0 m/s and at an angle of 40.0 degrees above the horizontal. The wall is 22.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? (b) What are the horizontal and vertical components of its velocity as it hits the wall? (c) When it hits, has it passed the highest point on its trajectory?"

    A friend told me to use Vy=25sin(40) = 16.07 but I'm not sure what that's for.

    I found that the ball hits the wall after .88 seconds, and I think the height of the wall might be 18.46 m but I'm not sure.

    I think I'm just confused about equations, vertical/horizontal components, and magnitude. Okay that's actually alot :[

    Thank you!
     
    Last edited: Oct 15, 2006
  2. jcsd
  3. Oct 15, 2006 #2
    assuming there is no air resistance, calculate the time needed for the ball to reach the wall, using the horizontal component of the initial velocity. The time taken shouldnt be .88s, recheck ur answer. Then, u can check the height at which the ball collides with the wall by using the vertical component of velocity.

    if there is no air resistance, horizontal component of the velocity shouldnt change, ya? To find the vertical velocity, use the formulas u learn from this topic. Ur acceleration should be simply be the acceleration of freefall.


    You should be able to reach a conclusion for this from ur ans from part b
     
    Last edited: Oct 15, 2006
  4. Oct 15, 2006 #3
    I see how you calculated the .88 seconds. It's incorrect. You'll need to study how to break a vector into perpendicular components. The motion of a ball fired at 40.0 degrees can be studied by observing the vertical motion of the ball independently from the horizontal motion of the ball.

    You should, by now, be aware that a cannonball fired horizontally from the top of a cliff takes the same amount of time to reach the ground as a ball simply dropped from the top of the cliff. The horizontal motion is independent of the vertical motion. The vertical motion will determine how much time it's in the air.

    Perhaps a good starting point is to look at displacements. Let's say Bob walks 3 meters East, then turns and walks 4 meters North. He's now 5 meters away from where he started (pythagorean theorem) at an angle of (whatever the solution to tan(theta)=4/3; 53ish degrees)

    Now, instead, I could say that Bob walked 5 meters at an angle of 53ish degrees. You'd have to work this problem backwards using basic trig to figure out how far East he is, and how far North he is. (Draw a right triangle.)

    Velocity is a vector, just like displacement. So, if instead of Bob walking, I said that we fired a cannonball 5 meters per second at an angle of 53 degrees, you could use a right triangle and figure out that the ball is moving horizontally at 3 meters per second, and it's initial velocity in the vertical direction is 4 meters per second. There are no forces in the x-direction; the velocity stays at a constant 3 m/s (ignoring air resistance.) But, in the y-direction, the velocity in the y direction is going to change due to the acceleration of gravity. Since the velocity in the y direction starts at 4 m/s, it will take 4m/s divided by 9.81 m/s^2 (roughly .4 seconds) for its vertical velocity to reach zero. This is the highest point (vertex) of the parabolic trajectory.

    Hopefully this helps... once you understand, the problem becomes quite simple.
     
  5. Oct 15, 2006 #4
    When I quickly did the calculation given the problem, it took 1.15s to reach the wall at the height of 24.97m.

    As gunblaze says, the horizontal component of the initial velocity shouldn't change (therefore should stay at 25*cos(40))

    Your friend is right about the vertical component of the initial velocity being 25*sin(40). Basically this is because the initial velocity is at the "diagonal", so you have to break this into x and y components using vectors (draw a little diagram, it always helps).

    I hope it helped!
     
  6. Oct 15, 2006 #5
    Wow, thanks for all the help!
    I'm still confused though.. :[

    Is the horizontal component 19.15?
    I'm still not sure how to find the seconds it takes for the ball to reach the wall.. what equation am I supposed to use?

    I'm not sure why I can't get this.. ahh :[
     
  7. Oct 15, 2006 #6
    Yup. The horizontal component is 19.15. You got that right. Now, you were told that the distance of the wall from the point of release is 22m right? So how u find time given distance and speed?
     
  8. Oct 15, 2006 #7
    So I use dx=vxt and get 22=19.5t and then t=1.15s!

    And then, the vertical component is 16.07-9.8(1.15) = 4.8? That doesn't seem right to me..
     
  9. Oct 15, 2006 #8
    Alright, now u got the time, what next?

    You know that the vertical component of ur initial velocity is 25sin40 right?
    From projectile motion, you should know the formula S=ut+1/2at^2. You got ur t which u had already found and u got the initial vertical velocity. You also know that without air resistance, ur acceleration is equal to ur g value right? so can u find ur displacement now?
     
  10. Oct 15, 2006 #9
    OH so i do S=(16.07)(1.15)+(1/2)(9.8)(1.15)^2 = 24.96!!

    wow i think thats it! thank you!
    the only other thing is.. why is the g value not negative?
     
  11. Oct 15, 2006 #10
    g value that you used in the calculation is -9.8. If you have used +9.8, you wouldn't get 24.96

    You must've just made a typo! Good job on getting the right answer though!

    EDIT: Never Mind... the correct equation that you must use is:
    [tex] s = v_{o}t - \frac{1}{2}at^{2} [/tex]
    But since a = g = -9.8, the negative sign just becomes positive.
     
    Last edited: Oct 16, 2006
  12. Oct 16, 2006 #11

    ohh okay that makes sense now.. for some reason I kept thinking it was an addition instead of subtraction. Thanks so much!
     
  13. Oct 16, 2006 #12
    ya. the g value must be a negative since it is acting downwards, in a direction opposite to the direction of the vertical component of the velocity
     
  14. Oct 16, 2006 #13
    Yeah I thought it had to be negative, I just had the equation mixed up.

    But how do I figure out if the ball has already made it to the highest point of its trajectory?
     
  15. Oct 16, 2006 #14
    What you can do is find an equation of velocity (I don't remember exactly what it is). Because at the top of the trajectory the vertical velocity is equal to zero, you can set one hand of the equation equal to zero, and solve for either s or t. From that information, you should be able to figure out it it passed the max or not (if t_max < t in part a, or s_max > s in part a, then it has already passed the max)
     
  16. Oct 16, 2006 #15
    oh okay.. im gonna figure out what equation to use then.. thanks :]
     
  17. Oct 16, 2006 #16
    What you can do is to find the vertical component of velocity as it hits the wall. Work out ur workings for part b first. From there, if ur vertical component of velocity is not equal to zero, +ve, then it has not yet reach its highest point of trajectory. If it is zero, then it has reached its point of trajectory. If it is -ve, it has passed its highest point of trajectory. To find the final vertical component velocity, you can use v^2=u^2+2as
     
    Last edited: Oct 16, 2006
  18. Oct 16, 2006 #17
    For v^2, I got -230.971 .. does that count as passing the highest point of trajectory?
     
  19. Oct 16, 2006 #18
    You know what? I'm an idiot. I'm going to scrap what I've said before and the answer that I've given you.

    If you think about it. If the vertical component of the velocity is 16.07, then it won't reach the height 24.97 even if there was no acceleration. The correct equation is:
    [tex] s = v_{o}t + \frac{1}{2}at^{2} [/tex]
    and you will need to put in g=-9.8, so it becomes
    s = 16.07(1.15) + 1/2(-9.8)(1.15)^2 = 12.0m

    I'm so sorry for the confusion, and yes, this is my final answer :redface:

    As far as the last question goes, you should get t=1.64 and therefore it hasn't passed the highest point yet.
     
  20. Oct 16, 2006 #19
    OHH. Well, that makes sense since that's what I got at first but thought 24.97 was the right answer.
    No problem, just thanks for helping me I have 10 more problems to do and its so hard for me :[

    Ok, thanks about the last question too!
     
  21. Oct 16, 2006 #20
    Yeah, I feel terrible. Sorry about that!

    For the last problem, the equation you should use is:
    [tex] v = v_{o} + at [/tex]
    Let v = 0, you know [tex] v_{o} [/tex] and you know a, so solve for t.

    You will get t = 1.64, and the ball hits the wall at t = 1.15, so it hasn't reached the maximum when it hits the wall.
     
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