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 17.3 m/s at an angle of 50.0˚ 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.
v^2 = v0^2 + 2ax
The Attempt at a Solution
Basically, what i did was getting the y value from the speed.
It was like: 17.3sin(50.0) = 13.25m/s. From the given, what
I could figure out were acceletion (-9.81m/s) and velocity (13.25m/s)
Since I only knew those two, I thought that I must use v^2 = v0^2 + 2ax.
For the maximum height, I did 0 = (13.25)^2 + 2(-9.81)(x) and x was 8.948m.
However, there was 3.00 elevation, so I subtracted 3.00 from 8.948,
which was 5.948m. From this, I tried to do the same equation and I
got the 10.80m/s for the answer, but the internet site says that it is wrong.