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 degree ramp, and 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 = 0.810 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/s^2. v_1 = m/s

Explanation / Answer

Suppose speed of ball is v as it leaves the ramp.

after that it will fly with 45 deg initiALLY.

So range have to be 2.20 m.

R = v^2 sin(2 x 45) / g

2.20 = v^2 / 9.81

v = 4.64 m/s

now applying energy conservation to find initial velocity of ball.

PEi + KEi = PEf + KEf

0 + m v0^2 /2 = m g h + (m v^2 /2 + I w^2 /2 )

for golf ball: I = 2 m r^2 / 5 and w = v/ r

v0^2 /2 = g h + v^2 /2 + v^2 / 5

v0^2 /2 = (9.81 x 0.810) + (7 x 4.64^2 / 10)

v0 = 6.79 m/s ..........Ans

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