a ball is thrown upward from an 80 ft. tower with an initial vertial speed of 40 ft/s.?

Question

Calculate the ball's speed just before it reaches the ground.
Thanks in advance!!!!

Answer 1

The potential energy of the ball when you throw it is m*g*h; the kinetic energy of the ball is .5*m*v^2. When the ball reaches the ground, all of this energy will be kinetic:
.5*m(vg)^2=m*32*80+.5*m*(40)^2.
You can simplify this a little by canceling out the m on both sides and getting rid of the .5: (vg)^2=5120+800=5920
Therefore, Vground = 76.9 ft/sec.