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 launching the ball into the hole which is d = 1.90 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 = 0.600 m, at what speed must the golfer hit the ball to land the ball in the hole? Assume a frictionless surface (so the ball slides without rotating). The acceleration due to gravity is 9.81 m/s^2.

Explanation / Answer

here,

let the speed must be the goalfer hit the ball be u0

let the speed at the height 0.6 m be u

using conservation of energy

0.5 * m*u0^2 - 0.5 * m*u^2 = m*g*0.6

u = sqrt( u0^2 - 11.76 )

luanch angle at the top , theta = 45 degree

horizontal distance d = 1.9 m

u^2 * sin(2 * theta) /g = 1.9

( u0^2 - 11.76 ) * 1 /9.8 = 1.9

u0 = 5.51 m/s

the speed must be the goalfer hit the ball is 5.51 m/s

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