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A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0 above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

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the ball reaches a maximum height of

h max = v0^2/2g with v0v = vertical component of velocity v0 = v0*sin40

hmax = (14*sin40)^2/(2*9.81)

hmax = 4.13 m

the ball has then to fall down by 4.13 - 3 m = 1.13 m to hit the ground.

Its vertical velocity is then vv = ?(2gh)

vv = ?(2*9.81*1.13 =4.70 m/s

and its horizontal velocity is v0*cos40= 14cos40= 10.72m/s

The total velocity is then ?(4.7^2+10.72^2) = 11.70 m/s <--- answer

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