Question

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after being released?

A. 49 joules
B. 98 joules
C. 1.0 × 10^2 joules
D. 1.1 × 10^2 joules

Answer 1

A. 49 joules

Step By Step Solution:

Via kinematic equations we can must first determine the velocity of this object after 10 meters (since halfway of 20 meters is 10).

`v_(final)^2=v_(initial)^2+2gd`

Where the initial velocity is zero since it's held at a height of 20 meters, hence it's not moving initially, g is the average gravity on Earth 9.8 m/s^2 and d is the distance this object will travel so 10 meters. By plugging these values in we obtain:

`v_(final)^2=0+2(9.8)(10)\\\\v_(final)=√(2(9.8)(10))=14m/s`

And the equation for kinetic energy is :

`K.E.=(1)/(2)mv^2=(1)/(2)(0.5kg)(14m/s)^2=49J`

So the kinetic energy halfway after being released is 49 joules.