A 6.43 kg ball is thrown upward with an initial speed of 14.34 m/s. Ignore air r
ID: 2103363 • Letter: A
Question
A 6.43 kg ball is thrown upward with an initial speed of 14.34 m/s. Ignore air resistance when answering the following questions.
a) What is the ball's initial kinetic energy?
b) What is the ball's kinetic energy when it reaches its maximum height?
c) What is the ball's gravitational potential energy (relative to Earth's surface) when it reaches its maximum height?
d) What is the ball's maximum height?
e) What is the ball's kinetic energy just before it returns to your hand?
Please show all math, I want to understand how to do this correctly.
Explanation / Answer
6.43 kg ball is thrown upward with an initial speed of 14.34 m/s.
a) 1/2 * m*v^2 = 0.5 * 6.43 * 14.34^2 =661.1184 J
b)kinetic energy when it reaches its maximum height = 0J because velocity at max height is 0 and all KE is converted into PE by law of conservation of energy
c) ball's gravitational potential energy (relative to Earth's surface) when it reaches its maximum height = initial kinetic energY(BY LAW OF CONSERVATION OF energy ) = 661.1184J
d)ball's maximum height
mgh=661.1184 h = 661.1184/(9.81*6.43) = 10.480 metre
e)ball's kinetic energy just before it returns to your hand = is same as initial kineteic energy = 661.1184 J