Question

A baseball outfielder throws a baseball of mass 0.15 kg at a speed of 40 m/s and initial angle of 30 degrees. What is the kinetic energy of the baseball at the highest point of the trajectory? Ignore air friction.

a. zero
b. 30 J
c. 90 J
d. 120

Answer 1

Kinetic energy, E = 90 J

Step-by-step explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Speed of projection, v = 40 m/s

Initial angle with horizontal,
`\theta=30^(\circ)`

At the highest point, the vertical component of velocity is equal to 0. There is only horizontal component of velocity that is given by :

`v_x=v\ cos\theta`

`v_x=40\ cos(30)`

`v_x=34.64\ m/s`

Kinetic energy at the highest point the kinetic energy of the baseball is given by :

`E=(1)/(2)mv_x^2`

`E=(1)/(2)* 0.15\ kg* (34.64\ m/s)^2`

E = 89.99 J

or

E = 90 J

So, the kinetic energy of the baseball at the highest point of the trajectory is 90 J. Hence, the correct option is (c).