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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 17.4 m/s at an angle of 35.0¢ª 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.

I was going to do vx= 17.4 cos 35

vy=(17.4 sin 35) - 9.8(?)

and then do sqrt vx^2 + vy^2 but I wasn't sure if the vy equation was correct?

• Physics(Please respond) -

Vx remains constant. You are correct on its value.
Vy2 decreases to Vyo^2 - 2 g *(3 m)

Then take the sqrt of Vx^2. + Vy^2

• Physics(Please respond) -

so for vy I would do (17.4 sin 35) - 2(9.8)(3) ??

• Physics(Please respond) -

I meant (17 sin 35)^2

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