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+1 SCIENCE PHYSICS

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A FROG SAW AN INSECT ON THE GROUND AT A HORIZONTAL DISTANCE OF 30 METER FROM IT. THE FROG CAN JUMP WITH A SPEED 20 METER/SECOND IN ANY DIRECTION. IT JUMPS WITH HIS INITIAL SPEED AND LAND ON THE INSECT.

A: WHAT IS THA TYPE OF MOSSION MADE BY THE FROG?

B: FIND THE ANGLE THAT THE FGOGS INITIAL VELOCITY MAKES WITH THE HORIZONTAL,FIND THE MINIMUM DISTENCE THE INSECT HAVE 2 KEEP FROM THE FROG SO THAT IT CAN ESCAPE FROM THE FROG?

  • +1 SCIENCE PHYSICS - ,

    The path of the frog is parabolic

    u = 20 cos A = constant

    d = range = u t = 20 cos A t = 30

    so t cos A = 1.5 and t = 1.5/cos A

    vi = 20 sin A

    v = 20 sin A - 9.8 t

    h = 0 + 20 t sin A - 4.9 t^2
    h = 0 when frog hits bug
    0 = t (20 sin A - 4.9 t)
    so t = 4.08 sin A

    so
    4.08 sin A = 1.5/cos A
    and
    sin A cos A = .3675

    sin A (1-sin^2A)^.5 = .3675
    (1-sin^2 A)^.5 = .3675/sin A
    1 - sin^2 A = .135/sin^2 A
    sin^2 A - sin^4 A = .135

    sin^4 A - sin^2 A + .135 = 0
    let x = sin^2 A

    x^2 - x + .135 = 0
    x = [ 1 +/-sqrt (1-.54) ]/2
    x = [ 1 +/- .678 ]/ 2
    x = sin^2 A = .322 or .839
    sin A = .567 or .916
    A = 34.5 deg or 66.3 deg

    max range when A = 45 deg
    u = 20 cos 45 = 14.1
    vi = 20 sin 45 = 14.1

    time in air
    v = vi - g t
    top when v = 0 halfway through flight
    t at top = 14.1/9.8 = 1.44
    so time in air = 2.88 seconds
    d = u t = 14.1 * 2.88 = 40.7 meters

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