An outfielder throws a 0.150-kg baseball at a speed of 56.0 m/s and an initial angle of 30.0°. What is the kinetic energy of the ball at the highest point of its motion?

Answer :

Kinetic energy of the ball at the highest point of its motion is 176.42 J

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

Mass, m = 0.150 kg

At highest point the baseball is only having horizontal velocity.

Horizontal velocity, v = 56 cos 30 = 48.50 m/s

Substituting

Kinetic energy = 0.5 x Mass x Velocity²

Kinetic energy = 0.5 x 0.15 x 48.50²

Kinetic energy = 176.42 J

Kinetic energy of the ball at the highest point of its motion is 176.42 J

Lanuel

The kinetic energy of the ball at the highest point of its motion is equal to 176.42 Joules.

Given the following data:

  • Mass of baseball = 0.150 kg
  • Speed of baseball = 56.0 m/s

To calculate the kinetic energy of the ball at the highest point of its motion:

The horizontal velocity of an object.

The baseball has a horizontal velocity at the highest point of its motion, which is given by this:

[tex]V_y=Vcos\theta\\\\V_y=56.0 \times cos30\\\\V_y=56.0 \times 0.8660\\\\V_y=48.50\;m/s[/tex]

The formula for kinetic energy.

Mathematically, kinetic energy (K.E) is given by this formula;

[tex]K.E = \frac{1}{2} MV^2[/tex]

Where:

  • K.E is the kinetic energy.
  • M is the mass of an object.
  • V is the velocity of an object.

Substituting the given parameters into the formula, we have;

[tex]K.E = \frac{1}{2} \times 0.150 \times 48.50^2\\\\K.E =0.075 \times 2352.25[/tex]

K.E = 176.42 Joules.

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