Results 1 to 4 of 4

Math Help - Initial speed and time

  1. #1
    Newbie
    Joined
    Aug 2010
    Posts
    8

    Initial speed and time

    OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.

    The equations I am using are time = square root(2y/-g), and initial velocity = x/t.


    For time I get square root((2*(-10m))/9.8m/s^2) = 1.43 seconds.
    For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.

    However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member BAdhi's Avatar
    Joined
    Oct 2010
    From
    Gampaha, Sri Lanka
    Posts
    252
    Thanks
    6
    you cannot apply S=ut when there is acceleration. So to find initial velocity, apply S=ut+\frac{1}{2}at^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by learningguy View Post
    OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.

    The equations I am using are time = square root(2y/-g), and initial velocity = x/t.


    For time I get square root((2*(-10m))/9.8m/s^2) = 1.43 seconds.
    For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.


    However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?

    y_1(t)=20-\frac{g}{2}t^2 and
    y_2(t)=v_0t-\frac{g}{2}t^2

    Using the first equation gives

    \displaystyle 10=20-\frac{g}{2}t^2 \iff t=\pm 2\sqrt{\frac{5}{g}}

    putting this into the 2nd equation gives

    \displaystyle 10=v_0\left( 2\sqrt{\frac{5}{g}}\right)-\frac{g}{2}\left( 2\sqrt{\frac{5}{g}}\right)^2 \implies v_0=2\sqrt{5g}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2009
    Posts
    162
    Quote Originally Posted by learningguy View Post
    OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.

    The equations I am using are time = square root(2y/-g), and initial velocity = x/t.


    For time I get square root((2*(-10m))/9.8m/s^2) = 1.43 seconds.
    For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.

    However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?
    Your time is correct. However, your other equation is wrong because there is a constant, acceleration due to gravity, in place (assuming it is on Earth). As a result you have to use the Δy=viΔt + 1/2a(Δt)^2 formula. Then you have to re-arrange it to solve for vi. Remember, to use your vectors correctly.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find the minimum initial speed....
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 29th 2010, 06:03 AM
  2. Initial speed of object sliding to a stop w/ air resistance
    Posted in the Differential Equations Forum
    Replies: 12
    Last Post: September 6th 2010, 03:32 AM
  3. Replies: 10
    Last Post: February 20th 2010, 03:34 PM
  4. Replies: 1
    Last Post: September 30th 2008, 02:10 PM
  5. find initial speed
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: September 16th 2008, 05:36 PM

Search Tags


/mathhelpforum @mathhelpforum