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: Check my numbers?

  1. Mar 24, 2004 #1
    Sorry, this is extremely basic for most of you so thanks for even reading this...

    "Eric is running to school and leaping over puddles as he goes. From the edge of a puddle 1.5 m long, he jumps .2 m high off the ground with a horizontal velocity of 3 m/s. WIll he land in the puddle?"

    I said: .2=4.9t^2
    .04...=t^2, t=0.2 s
    0.2*3=0.6, 0.6 < 1.5 therefore, yes he will hit the puddle.
     
  2. jcsd
  3. Mar 24, 2004 #2
    The total time of the jump is actually [tex]2t[/tex], so the horizontal distance is 1.2 meters. Still, he won't make it over the paddle.
     
  4. Mar 24, 2004 #3
    Thanks Chen, I almost forgot t is jsut for one direction.

    I have a question on this one too:

    "A golf ball is hit, it travels 155 yards before hitting the ground 3.2 seconds later. What was it's initial velocity?"

    Now, I found the height to be about 12.5 yards, but I don't know where to go from there.
     
  5. Mar 24, 2004 #4

    ShawnD

    User Avatar
    Science Advisor

    Just make 2 simultaneous equations.
    I'll write the distances x3 (3 feet in a yard i think).

    horizontal distance (solve for velocity):

    [tex]d_x = V_xt[/tex]

    [tex]d_x = Vcos(\theta)t[/tex]

    [tex]V = \frac{d_x}{tcos(\theta)}[/tex]

    vertical distance (solve for velocity):

    [tex]d_y = V_yt + \frac{1}{2}at^2[/tex]

    [tex]0 = Vsin(\theta)(t) + \frac{1}{2}at^2[/tex]

    [tex]V = \frac{-at}{2sin(\theta)}[/tex]

    Now if I did those right, this should work. Make the equations equal to each other and try to solve for theta (the angle)

    [tex]V = V[/tex]

    [tex]\frac{d_x}{tcos(\theta)} = \frac{-at}{2sin(\theta)}[/tex]

    [tex]tan(\theta) = \frac{-at^2}{2d_x}[/tex]

    [tex]tan(\theta) = \frac{-(-32.2)(3.2)^2}{2(465)}[/tex]

    [tex]\theta = 19.52[/tex]

    Now back to the first equation

    [tex]V = \frac{d_x}{tcos(\theta)}[/tex]

    [tex]V = \frac{465}{(3.2)cos(19.52)}[/tex]

    [tex]V = 154.17 \frac{ft}{s}[/tex]


    So the answer would be 154.17 ft/s at 19.52 degrees from the horizontal.
    Remember how to solve for simultaneous equations, it will come back to haunt you.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook