golf tees
goaliejoe asked:


A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 44 m/s at an angle of 26° above the horizontal. The ball strikes the fairway a horizontal distance of 200 m from the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?

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  • Comments

    (Ω)Mistress Bekki on 11 February, 2010 at 5:46 am #

    initial vertical speed = vy = v sin theta
    initial horizontal speed = vx = v cos theta

    range = flight time * horizontal speed
    r = t * vx

    Solve for the flight time
    t = r / vx = r / (v cos theta)

    The height as a function of time is:
    h(t) = vy t - 1/2 gt^2

    Plugging in our flight time:

    h = v sin (theta) r / (v cos theta) - 1/2 g (r / (v cos theta))^2

    = r tan (theta) - g r^2 / (2 v^2 cos(theta)^2)

    b) Use conservation of energy
    initial KE = 1/2 mv^2
    =
    final KE + PE = 1/2 m vf^2 + mgh

    Solve for your landing speed:
    vf = sqrt (v^2 - 2gh)
    = sqrt (v^2 - 2 g (r tan (theta) - g r^2 / (2 v^2 cos(theta)^2))

    Have fun plugging in numbers.