kinetic energy of the baseball at the highest point of the trajectory?

Question

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

Answer 1

The horizontal component of speed
= 40 m/s * cos(30)
=40 cos(30) m/s
mass
=0.15 kg
Since at the highest point, vertical velocity=0, and horizontal velocity is unchanged (ignore air resistance), so
K.E.=mv^2/2=0.15*(40cos(30))^2/2
=0.15*1600*3/4/2
=90 joules

Answer 2

At the time the outfielder threw the ball, the velocity had a vertical and horizontal component. But at the highest point the vertical component of the velocity became equal to zero. What remained was the horizontal component of the velocity which is constant during the flight of the ball.

Kinetic Energy, KE=1/2mv^2, where v in this formula is the horizontal velocity equal to 40cos30 or 34.6m/s.

Now solve for KE by substituting known values:

KE=1/2*0.15*34.6^2
=89.8 Joules

Answer 3

In a trajectory, at the highest point, VERTICAL velocity (vy) is zero. At all points in the trajectory (start, highest point, end, everything in between), the horizontal component of the velocity (vx) is constant.

So, at the highest point in the trajectory, use KE = 1/2 m(v^2), but v is the velocity in the x-direction (vy is zero)

Answer 4

This one's easy. At the peak of its trajectory, it possesses only a horizontal component to its velocity, which is the same as its initial horizontal velocity. Vx = V[cos(theta)] = 40*0.866 = 34.6 m/s. KE at its highest point = 0.5*m*v^2 = 0.5*0.15*34.6^2 = 90 J.

Answer 5

superb question.... The balls trajectory could be divided right into a x (distance) and y( height) element. The y speed is lowest on the splendid of the trajectory. The x speed is 0 on the tip factor. So i'm afraid it rather relies upon on the attitude the ball is struck.

Answer 6

vertical velocity=0
horizontal velocity=40*cos(30)
Kinetic energy=mv^2/2= 0.15*(40*cos(30))^2/2 =90 Joule

Answer 7

pretty fast