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Projectile motion question need help

  1. Oct 8, 2012 #1
    A golfer drives a golf ball 310.24 m down the fairway. If the ball is launched at an angle of 24.5 o to the horizontal, what is the maximum height attained by the ball during its flight?


    H=1/2gt^2
     
  2. jcsd
  3. Mar 12, 2013 #2
    The maximum height attained is given by [tex] H = \frac{u^2 \sin^2 \theta}{2 g} [/tex] where u is the magnitude of the initial velocity provided and θ the angle with the horizontal.

    The range of the projectile is [tex] R = \frac{u^2 \sin 2 \theta}{g} [/tex] which is given to be 310.24 m.

    Thus we have [tex] H = \frac{R}{4} \tan \theta [/tex]
     
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