Question

A 47 gram golf ball is driven from the tee with an initial speed of 52 m/sec and rises to a height of 24.6 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8 m below its highest pint

Answer 1

a) 52.2 J

b) 48.77 m/s

Step-by-step explanation:

a)47 g = 0.047 jg

The kinetic (and total mechanical energy) of the ball at the ground is

`E = mv^2/2 = 63.544 J`

The potential energy of the ball at its highest point 24.6m is. Let g = 9.8m/s2

`E_h = mgh = 0.047*9.8*24.6 = 11.33 J`

Since the potential energy at the highest height is less than the total mechanical energy on ground, the difference must be kinetic energy

`E_k = E - E_h = 63.544 - 11.33 = 52.2J`

b) 8m below 24.6m is 16.6m. The potential energy at this point is

`E_(p8) = mgh = 0.047*9.8*16.6 = 7.64 J`

And so the kinetic energy at this point is

`E_(k8) = E - E_(p8) = 63.544 - 7.64 = 55.9 J`

So the speed is

`mv^2/2 = 55.9`

`v^2 = 2*55.9/0.047 = 2378.64`

`v = √(2378.64) = 48.77 m/s`