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 launching the ball into the hole which is d = 2.20 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 = 1.000 m, at what speed must the golfer hit the ball to land the ball in the hole? Assume a frictionless surface and the acceleration due to gravity is 9.81 m/s2.

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 launching the ball into the hole which is d = 2.20 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 = 1.000 m, at what speed must the golfer hit the ball to land the ball in the hole? Assume a frictionless surface and the acceleration due to gravity is 9.81 m/s2.

Explanation / Answer

After leaving the tamp ...ball travels in projectile motion let it has a velocity v when leaving the ramp top horizontal component = vcos(45) time of flight T= 2vsin(45)/g 2.2 = vcos(45)*T = v^2sin(90)/g = v^2/g v^2 = 21.56 kinetic energy at the top of ramp = .5mv^2 u = initial speed initial kinetic energy = .5mu^2 Potential energy gain when it is at top of the ramp = mgy = 9.81m energy conservation .5mu^2 -.5mv^2 = mgy u^2 - v^2 = 2*9.81 u (v1)= 6.42 m/s I hope this is clear :)

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