A golf ball is hit at point A with velocity Va=40 m/s directed at a 40 degree an

Question

A golf ball is hit at point A with velocity Va=40 m/s directed at a 40 degree angle with the horizontal as shown. Find the distance d where the ball strikes the flat 10 degree inclined surface at point B.

The answer is 128.8 meters. Thanks in advance!

Explanation / Answer

Resolving the coordinate system along the inclined plane and perpendicular to the inclined plane: velocity perpendicular to the inclined plane = 40 sin(40) net displacement perpendicular to the inclined plane = 0 acceleration perpendicular to the inclined plane = - (9.8)cos(10) => 0 = 40sin(40)t - (0.5)((9.8)cos(10))(t^2) => t = 40sin(40)/[(0.5)((9.8)cos(10))] => t = 5.3282 seconds Now, initially velocity along the inclined = 40 cos(40) acceleration along the incline = -9.8sin(10) => displacement along the incline = (40cos(40))(5.3282) - (0.5)(9.8sin(10))(5.3282)^2 => displacement along the incline = 139.12 the answer should be 139.12 metres

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