Kmol6 said:
The velocity of the x and y components are 175 (cos 45) = 123.7 and 175(sin 45) = 123.7
The ball is initially at 165m above the cliff, but projects up further ( i don't know how high though?) at that point the potential energy is equal to the final kinetic energy when the ball hits the ground and at that point the balls potential energy is 0.
but I'm still confused as to what I put into the equation? Is it just a 1 step or 2 and if so how do I figure out the height of the projectile using energy?
What is the magnitude of the velocity? Does the launch angle matter?
That is not correct and this is an important point to make. At the peak of its trajectory, the potential energy is
not equal to the final kinetic energy. For that to be true, the ball's kinetic energy would have to be zero, but if you remember what the trajectory of a projectile looks like, it's a parabola, not a vertical line, meaning the ball is still moving forward, so its total kinetic energy can't be zero. At its highest point, its
vertical velocity is zero (such that it won't go up any higher), but it's horizontal velocity is still the same, so the ball does indeed have kinetic energy at that point, though less kinetic energy than it initially had.
What the conservation of energy tells is that the sum of the kinetic and potential energy is going to be constant. That means it doesn't matter which point of flight we choose to measure them. The value of their sum is always going to be the same. To make our lives easier, we can find the sum at the time when it is initially launched because we can easily find both the kinetic and potential energy at that time. That sum must then be equal to its final kinetic energy.
As you said, the final potential energy is equal to zero and so the equation becomes
\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2
If we take its initial velocity to be its launch velocity and its initial potential energy to be the potential energy at its initial height? What values will you use?
