Asked by Dot

A golfer hits a shot to a green that is elevated 3.00 m above the point where the ball is struck. The ball leaves the club at a speed of 17.8 m/s at an angle of 53.0˚ 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.
Answered by alisha
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Answered by Steve
h(t) = 3.00 + 17.8 sin53.0˚ - 4.9t^2
find t when h=0

Then, recall that the upward speed is

v<sub>y</sub> = 17.8 sin53.0˚ - 9.8t
v<sub>x</sub> = 17.8 cos53.0˚

the final v is of course found by

v^2 = v<sub>x</sub>^2 + v<sub>y</sub>^2
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