Question

The center fielder for the school's baseball team throws a 0.150-kg baseball towards home plate at a speed of 22.0 m/s and an initial angle of 40.0°. what is the kinetic energy of the baseball at the highest point of its trajectory?

Answer 1

at the highest point of the trajectory of the ball only horizontal component of the velocity will remain as there is no acceleration in horizontal direction.

But due to gravity in vertical direction the vertical component of the velocity will become zero at the highest point

so here we can say that at highest point we will have

$v_x = 22 cos40 = 16.85 m/s$

$v_y = 0$

now for the kinetic energy we can say

$K = \frac{1}{2} mv^2$

here we know that

$m = 0.150 kg$

$v = v_x = 16.85 m/s$

now the kinetic energy will be given as

$K = \frac{1}{2}*0.150*16.85^2$

$K = 21.3 J$

so kinetic energy will be 21.3 J