1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Projectile Motion ball is launched

  1. Nov 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A ball is launched with initial speed ##2\sqrt{}gh## such that it just clears two walls of height 'h' separated by a distance 2h .What is the time taken by the ball between the two walls ?

    Ans: ##2\sqrt{\frac{h}{g}}##

    2. Relevant equations

    3. The attempt at a solution

    Considering the origin to be at the launching point ,θ to be angle at which the ball is projected and 'x' to be distance between the origin and the first wall .

    $$h = xtanθ-\frac{1}{2}\frac{gx^2}{u^2cos^2θ}$$

    $$h = (x+2h)tanθ-\frac{1}{2}\frac{g(x+2h)^2}{u^2cos^2θ}$$

    Solving the above two equations , we get x = h(2sin2θ-1) .

    Now sure what to do next . I would be grateful if somebody could help me with the problem.
    Last edited: Nov 19, 2014
  2. jcsd
  3. Nov 19, 2014 #2


    User Avatar
    Homework Helper

    ##2\sqrt{gh}## can not be angle. Is it initial speed?
  4. Nov 19, 2014 #3
    Sorry ... You are right . It is initial speed . I have corrected it.
  5. Nov 19, 2014 #4


    User Avatar
    Homework Helper

    The answer is also wrong.

    Anyway: you need time, so use the equations which include time.
    Last edited: Nov 19, 2014
  6. Nov 19, 2014 #5
    I have corrected the answer . Will reattempt this problem .

    Thank you very much .
  7. Nov 20, 2014 #6
    You do not care about the launching point and the initial launch angle. You can trivially the speed of the ball as it reaches the first wall, which reduces the problem to motion between the walls.
  8. Nov 20, 2014 #7
    Yes, shouldn't it be √(h/g)?
  9. Nov 20, 2014 #8


    User Avatar
    Homework Helper

    The OP has been corrected. The answer 2√(h/g) is right.
  10. Nov 20, 2014 #9
    Can you elaborate on the triviality of the speed of the ball as it reaches the first wall? I'm guessing it falls out of the symmetry of the problem, but I must say that it isn't obvious to me.
  11. Nov 20, 2014 #10
    The initial speed is known;
    so the initial kinetic energy known;
    the height of the wall is known;
    so the potential energy at the wall is known;
  12. Nov 20, 2014 #11
    Hah, fair enough, I was completely ignoring energy considerations.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted