Question

A fielder tosses a 0.15 kg baseball at 26 m/s at a 36 ∘ angle to the horizontal.Part AWhat is the ball's kinetic energy at the start of its motion?Express your answer with the appropriate units.Part BWhat is the kinetic energy at the highest point of its arc?Express your answer with the appropriate units.

Answer 1

Answer:

A. 50.7 J

B. 33.18 J

Explanation:

At the start of the motion the kinetic energy of the ball would be

$E_k = \frac{mv^2}{2} = \frac{0.15*26^2}{2} = 50.7  J$

When the ball gets to the highest point, vertical velocity must be 0, only horizontal velocity contributes to the kinetic energy. Since air resistant can be neglected, horizontal energy is preserved since the start

$v_h = vcos(36^o) = 26*0.81 = 21.03 m/s$

So the kinetic energy at this point is

$E = \frac{mv_h^2}{2} = \frac{0.15*21.03^2}{2} = 33.18J$

Answer 2

Given the data from the question, the kinetic energy at the start and at the highest point are:

A. The kinetic energy at the start of the motion is 50.7 J

B. The kinetic energy at the highest point is 0 J

What is kinetic energy?

This is the energy possessed by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

• KE is the kinetic energy
• m is the mass
• v is the velocity

A. How to determine the KE at the start

• Mass (m) = 0.15 Kg
• Velocity (v) = 26 m/s

• Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.15 × 26²

KE = 50.7 J

B. How to determine the KE at the highest point

• Mass (m) = 0.15 Kg
• Velocity (v) = 0 m/s (at highest point)

• Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.15 × 0

KE = 0 J

Learn more about energy:

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