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

Answer:

Kinetic energy, E = 90 J

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 :

$E=\dfrac{1}{2}mv_x^2$

$E=\dfrac{1}{2}\times 0.15\ kg\times (34.64\ m/s)^2$

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).