Koshka
Lv 5
Koshka asked in Science & MathematicsPhysics · 1 decade ago

Repost: How long before the drat touches the ground?

It is common knowledge that drats just "ball-up" and jump whenever they hear a car horn.

A drat of mass M is standing on the top of a 45 meter building overlooking a street during rush hour traffic.

An impatient driver honks the horn.

The drat jumps at a 30 degree angle from the top and hits the ground 73 meters away from the building.

How long did the drat take to hit the ground?

This should not be too hard.

http://s5.tinypic.com/xfzgk.jpg

2 Answers

Relevance
  • Pneuma
    Lv 5
    1 decade ago
    Favorite Answer

    73 = v cos30º t

    0 = 45 + v sin30º t - (1/2) g t²

    ==> 0 = 45 + 73 tang30º - (1/2) g t² ===>

    t = √ {2(45+73 tang30º)/g} = √ 17.785 =4.21723 sec

    (g = 9.8 m/sec²)

    Hope it help. Have a nice day.

    Note: My English is very poor, so I don't know about drats and I am still wondering about the role of the mass M in this scenario. I do think that M is irrelevant to answer the question, unless drats need some time (function of their mass) before jumping. They may have panicked before jumping and it is plausible that this state depends on their weight. In that case we need to know the "panic state time period" as a function of mass M. This PSTP(M) ---or ∏(M) for short--- should be added to the former calculated figure.

    In support of this consideration we show here a drat-cat during its panic state before jumping:

    http://www.musicalstonight.org/images/dratcat.jpg

  • Alex
    Lv 4
    1 decade ago

    You can do this using the equations of motion, and creating a simultaneous equation with the vertical and horizontal components of motion.

    ---

    Horizontal:

    u = u cos(30)

    a = 0

    s = 73

    t = t

    Vertical:

    u = u sin(30)

    a = -9.8

    s = -45

    t = t

    s = ut + 0.5att

    73 = u cos(30) x t

    -45 = u sin(30) x t - 4.9tt

    u = 73/(t cos(30))

    -45 = (73 t sin(30)) / (t cos (30)) - 4.9tt

    -45 = (73 sin(30))/cos(30) - 4.9tt

    4.9tt = 73rt(3)/3 + 45

    t = 4.2 seconds

    Hope that helps!

    --

    Note that the mass has no effect in this scenario, as gravitational pull is unrelated to the mass of an object.

    Cheers.

    Note #2 - I consider it unlikely that this question includes a reaction time. If this was the case, it is likely more information would be given and/or other figures would be requested in the question.

    Cheers.

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