What is the final velocity of the golf ball before it lands?

AI Thread Summary
The discussion focuses on calculating the final velocity of a golf ball before it lands on an elevated green. The initial velocity is given as 17.1 m/s at a 52.0˚ angle, with the ball reaching a maximum height before descending. Participants point out errors in the calculations, particularly regarding the maximum height and the need to consider both the vertical and horizontal components of the velocity. The correct approach involves using the kinematic equations to find the final speed just before landing. Accurate calculations are essential for determining the ball's final velocity, which is influenced by both components.
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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 17.1 m/s at an angle of 52.0˚ 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.

Initial velocity = 17.1sin52=17.1 x 0.788
=13.47 m/s.

max height = 9.81

v^2=u^2 + 2as

v^2=0+2*9.81*6.25

v= 11.07 m/s

My answer is wrong.
 
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fedora said:
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 17.1 m/s at an angle of 52.0˚ 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.

Initial velocity = 17.1sin52=17.1 x 0.788
=13.47 m/s.

max height = 9.81

v^2=u^2 + 2as

v^2=0+2*9.81*6.25

v= 11.07 m/s

My answer is wrong.
looks like you calculated max height as 9.25 not 9.81.
An extra step, but you correctly got the y component of the velocity. What about the x component?
 
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