posted by Bob on .
A ball of mass m = 0.75 kg is thrown straight upward with an initial speed of 8.9 m/s. Plot the gravitational potential energy of the block from its launch height, y = 0, to the height y = 5.0 m. Let U = 0 correspond to y = 0. Determine the turning point (maximum height) of this mass.
looks like a straight line to me.
The key here, and you were not asked to do it, is plot also the kinetic energy on the same graph.
KE=1/2 m v^2=1/2 m (vi^2-2*9.8h)
the sum of GPE and KE should be constant on the graph as in
mgh+1/2 m (vi^2-19.6h)=
mgh+1/2 m*2gh+1/2 mvi^2=
1/2 mvi^2 amazing, a constant.
If a baseball is thrown 30 m/s backwards from a truck moving 50 m/s, how fast will the ball strike the glove of a ground-based catcher
30 m/s - 50 m/s= -20m/s
the backwards represents subtraction. so simply subtract 30 - 50 = -20