1. Oct 8, 2012 #1

   richardjoe05

   

   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

    

   richardjoe05, Oct 8, 2012
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3. Mar 12, 2013 #2

   vik10622

   

   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]

    

   vik10622, Mar 12, 2013

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