A 47.0 g golf ball is driven from the tee with an initial speed of 48.0 m/s and rises to a height of 24.8 m.
(a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.
(b) What is its speed when it is 10.0 m below its highest point?
not sure why b won't be given by vf = vo^2 + 2*g*(yo - yf)
Let upward be the positive y direction, and let the zero point for the gravitational potential energy be at the level of the tee. There are no losses due to friction so we know that
Now, the ball started with a speed equal to 48.0 m/s and at a height equal to 0 m. The maximum height h is given as 24.8 m, so the KE at that point is:
The answer is that it does. However the v's in this formula are only the y components of the velocities in question. So you can use this equation, but once you get the y component of the final velocity, you need to add it vectorally to the x component of the final velocity, which is a difficult thing to do as we haven't been given the angle ball was hit at! (But it is quite possible to do this problem, it just takes a while.)
Again, think of energy techniques. I'll use the same coordinate system for this part of the problem. (You can use a new one with +y upward and the 0 for GPE at the max height. Then your final y is -10 m.)
Again we have
(The 42.7211 J is from part a. It is the kinetic energy of the ball when it is at max height.)
I get (after a lot of simplification)
(This might seem to be large in comparison to the initial speed, but most of it comes from the horizontal component of the velocity, which is about 42 m/s.)
Note: By the way, if you read your textbook carefully you will note that we can apply . The thing to note is that this is a vector equation. The form of this equation for more than 1D motion is
where the s's are displacement vectors and the between the vectors is the "dot product" or "scalar product" depending on how you learned it. (And I have defined as is typically done.)