A seasoned mini golfer is trying to make par on a tricky hole #5. The golfer can

Question

A seasoned mini golfer is trying to make par on a tricky hole #5. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45° ramp, and launching the ball into the hole which is d = 1.75 m away from the end of the ramp. If the opening of the hole and the end of the ramp are at the same height, y = 0.670 m, at what speed must the golfer hit the ball to launch the ball over the moat so that it lands directly in the hole? Assume a frictionless surface (so the ball slides without rotating). The acceleration due to gravity is 9.81 m/s2.

Explanation / Answer

We are given the angle and we know the opposite of the triangle so we can calculate the adjacent

tan = opposite/adjacent

tan 45 = 0.670/x

x= 0.670 ( since tan 45 is 1 the opposite must equal the adjacent )

the range is 1.75 meters + 0.670 meters (which is the adjacent of the triangle + the distance from the end of the ramp )

R= (Vo2 sin2)/g

(1.75+0.670)=(Vo2 sin90)/9.81

2.42=Vo2/9.81

Multiply both sides by 9.81

23.74=Vo2

Take the square root of both sides

Vo= 4.87m/s

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