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 = 1.15 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 = 0.960 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.

Sopling Learning Map make par on a tricky hole #5. The golfer can complete the hole by hiting the bal from the flat secion it lays on, up a da1 15 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 096o m, at what speed must the golfer hit the bal to lanch the ball over the moal so that 45 hamp, and launching the ball into the hole which is It lands d acoeleration due to gravity is 981 m/s? e hole? Assume a frictionless surtface (so the bal slides without rotating) The 1.15 m Number ,- 16,07 m/s oscotm er

Explanation / Answer

From projectile motion we have:
x = vx t
y = vy t - g/2 t^2
Which means at the time the ball returns to its starting height it has traveled a horizontal distance d = 2 vx vy / g.

Since the ramp is at a 45°, vx = vy = v sin(45) and the range is d = v^2 / g.

The fact that the surface is frictionless means no energy was lost, so we can use conservation of energy to relate the ball's speed at the top of the ramp to its initial speed. Lack of friction also means the ball is sliding instead of rolling, so we don't have to include a term for rotational kinetic energy.
1/2 m vi^2 = mgh + 1/2 m v^2

Putting this together with the range calculated earlier:
1/2 m vi^2 = mgh + 1/2 mgd
The m's cancel out.
vi^2 = 2gh + gd

vi^2 = 2*9.81*0.960 + 9.81*1.15

vi = 5.487m/s

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