The golf ball is hit at A with a speed of va = 40 ft s and directed at an angle

Question

The golf ball is hit at A with a speed of va = 40 ft s and directed at an angle of 30 degree with the horizontal as shown. Determine the distance d where the ball strikes the slope at B.

Explanation / Answer

X-MOtion: VAx = 40 cos 30 = 34.64 m/s, xA=0, xB = 5d/v(52+1) = 0.9806d For a constant x-velocity, we have xB = xA + VAx t => 0.9806d = 0 + 34.64 t => t = 0.02831 d (1) y-motion: VAy = 40 sin 30 = 20 m/s, yA=0, xB = d/v(52+1) = 0.1961d, ay=-g =-9.81 m/s2 For a constant y-acceleration, we have yB = yA + VAy t +ayt2 => 0.1961d = 0 +20 t-9.81t2/2 => 4.905 t2-20t + 0.1961d =0 (2) Plugging (1) into (2) and simplifying, we have 0.0039303d2 - 0.37002d =0 Soving we have d =0 (discard, since d can not be zero) d = 94.1 m Ans.

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