Results 1 to 6 of 6

Math Help - find initial speed

  1. #1
    Member
    Joined
    Nov 2005
    Posts
    172

    find initial speed

    A basketball player who is 2.00 m tall is standing on the floor L = 13.0 m from the basket, as in the figure below. If he shoots the ball at a 35.0 angle with the horizontal, at what initial speed must he shoot the basketball so that it goes through the hoop without striking the backboard? The basket height is 3.05 m.
    so far i got:
     1.05 = V_o sin(35)(\frac{13}{V_o cos(35)}) + \frac{1}{2}(-9.8)(\frac{13}{V_o cos(35)})^2

    need help solving for V_o
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by viet View Post
    so far i got:
     1.05 = V_o sin(35)(\frac{13}{V_o cos(35)}) + \frac{1}{2}(-9.8)(\frac{13}{V_o cos(35)})^2

    need help solving for V_o
    Cancel V_0 in the first term on the right hand side. Note that \frac{\sin 35^0}{\cos 35^0} = \tan 35^0, which simplifies that term even more. Now make V_0^2 the subject ....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2005
    Posts
    172
    after cancelling V_o i got
     1.05 = tan(35)(13) - 4.9 (\frac{13}{V_o cos35})^2

    my algebra is a little rusty, can you show me how to make V_o^2 the subject?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by viet View Post
    after cancelling V_o i got
     1.05 = tan(35)(13) - 4.9 (\frac{13}{V_o cos35})^2

    my algebra is a little rusty, can you show me how to make V_o^2 the subject?
    First isolate the second term by shifting all the constants to the left hand side.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2005
    Posts
    172
    let me know if im doing this wrong:
    1.05 = 9.10-4.9(\frac{13}{V_o cos(35)})^2

    -8.05 = -4.9(\frac{13}{V_o cos(35)})^2

    1.64 = \frac{13^2}{(V_o cos(35))^2}

    1.64 = \frac{168}{(V_o cos(35))^2}

     V_o^2 cos(35)^2+1.64 = 169

     V_o^2 cos(35) = 167.36

     V_o^2 = 204.309

     V_o = \sqrt{204.309} = 14.29
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by viet View Post
    let me know if im doing this wrong:
    1.05 = 9.10-4.9(\frac{13}{V_o cos(35)})^2

    -8.05 = -4.9(\frac{13}{V_o cos(35)})^2

    1.64 = \frac{13^2}{(V_o cos(35))^2}

    1.64 = \frac{168}{(V_o cos(35))^2} Mr F says: The 168 should be 169. Typo ....?

     V_o^2 cos(35)^2+1.64 = 169 Mr F says: Should be  {\color{red}V_o^2 cos(35)^2 \times 1.64 = 169} . There will be a ripple effect - you'll need to make the necessary corrections below.

     V_o^2 cos(35) = 167.36

     V_o^2 = 204.309

     V_o = \sqrt{204.309} = 14.29
    ..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Initial speed and time
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: February 6th 2011, 07:17 PM
  2. find the minimum initial speed....
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 29th 2010, 07:03 AM
  3. Initial speed of object sliding to a stop w/ air resistance
    Posted in the Differential Equations Forum
    Replies: 12
    Last Post: September 6th 2010, 04:32 AM
  4. Replies: 1
    Last Post: September 30th 2008, 03:10 PM
  5. Find the initial investment?
    Posted in the Business Math Forum
    Replies: 3
    Last Post: July 11th 2006, 08:59 AM

Search Tags


/mathhelpforum @mathhelpforum