K.E. = (1/2)M V^2 = 450 Joules
You can find that equation in my previous answer.
The Potential Energy at the top of the throw will be the same (450 J), since the velocity will be zero there, and the sum of Potential and Kinetic Energy does not change.
For the height H that is reached, solve
M g H = 450 joules
H = 450/(M*g) = 450/(1*9.8) = 45.9 m
What is the kinetic energy of a 1 kilogeram ball is thrown into the air with a velocity of 30m/sec?
How much potential energy does the ball have when it reaches the top of its ascent?
How high into the air did the nall travel?
Please help me help my child with her homework?
2 answers
Thank You for your help. It was very helpful. Now, I think I can get staarted with heling my child.