1. Not finding help here? Sign up for a free 30min 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!

Ball thrown from cliff

  1. Apr 3, 2005 #1
    Q. A man on the edge of a cliff H = 40 m high throws a ball directly upward. It returns past him 1.8 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)

    a) With what initial speed did the man throw the ball?

    Ans. H = 40m

    t = 1.8 seconds

    Using,

    x-x0 = v0t + 1/2 at^2

    To find vo:


    x -xo = 0 ( I wonder why x -xo , is zero,) when the height of the cliff is given to be 40m.

    0 = 1.8u + 4.9 * (1.8)^2

    Solving for u, we get:

    v = 8.82 m/s, which is correct I think!

    b) How high above the point of release did the ball go?

    We can use,

    v^2 = v0^2 + 2g (x-xo)

    v = final velocity = 0 m/s
    v0 from part a, is 8.82 m/s

    and g = 9.81 m/s2

    Plugging in all that, we get:

    x-x0 = 3.964 m

    c) What is the speed of the ball when it passes him on the way back down?

    Same as a) 8.82 m/s

    d) How long after the man throws it does the ball hit the ground at the bottom of the cliff?

    This is where I am stuck......

    Which equation to use?

    v^2 = u^2 + 2g ( x - xo)

    or

    x - xo = v0t + 1/2 gt^2

    and what would my variables be

    v0 = 8.82 m/s

    g = 9.81 m/s2

    We need to find 't' and add it to the 1.7seconds in the problem

    That would be the total time, to hit the ground,


    Plz. help!!! :cry:
     
  2. jcsd
  3. Apr 3, 2005 #2
    A man on the edge of a cliff H = 40 m high throws a ball directly upward. It returns past him 1.8 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.)

    a)With what initial speed did the man throw the ball? Correct

    The ball underwent constant acceleration for time = 1.8s.

    [tex] x-x_0 = v_0t + 1/2 gt^2 [/tex]
    [tex] x-x_0 = v_0(1.8) + 9.8*1.8^2/2 [/tex]

    x-x_0 = 0 since there is no total displacement, this is because the signs of v_0*t and 1/2 gt^2 are opposite.

    [tex] 0 = v_0(1.8) - 15.876 [/tex]
    [tex] v_0 = 15.876/1.8 = 8.82m/s [/tex]

    b) How high above the point of release did the ball go? Correct

    v_0 = 8.82
    h = h
    g = 9.8

    [tex] h = \frac{v_0^2}{2g} = \frac{77.79}{19.6} = 3.969m [/tex]

    c) Correct
    d) How long after the man throws it does the ball hit the ground at the bottom of the cliff?
    Theres two ways you can do this, you can take it as freefall from the point where its right infront of him, or freefall from the highest point. The second one is much easier and we'll do that, the ball goes 3.969m above where he threw it, for a total height of 43.969m from the ground.

    [tex] h = v_0t + \frac{gt^2}{2} [/tex]

    v_0t = 0 because at the top of the throw the ball is not moving.
    [tex] h = \frac{gt^2}{2} [/tex]

    From here we solve for time:

    [tex] t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{87.938}{9.8}} = 2.995s[/tex] plus the time for it to get from his hand to the peak = 0.9s.

    Total time = 3.89s

    Method 2:

    [tex] h = v_0t + \frac{gt^2}{2} [/tex]

    Solved for time via quadratic formula:

    [tex] t = \frac{2v+2\sqrt{v^2+2gh}}{2g} [/tex]

    [tex] t = \frac{2(8.82) + 2\sqrt{8.82^2 + 2(9.8)(40)}}{2(9.8)} = 3.895s[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Ball thrown from cliff
  1. Ball Thrown From Cliff (Replies: 1)

  2. Ball thrown to a cliff (Replies: 3)

Loading...