Question

A boy thrown a ball a mass 300gm with a velocity 25 m/s at an angle of 40 degree with the horizontal what is Kinetic energy of ball and heighest point on the trajectory​

Answer 1

$\huge\fcolorbox{black}{lime}{❥ᴀ᭄ɴsᴡᴇʀ}$

Given

Mass = 300 g

Velocity = 25 m/s

θ = 40°

To Find

Kinetic Energy of the ball at the heighest point of trajectory

Solution

☯ Kinetic Energy = ½mv²

☯ v = u cosθ

Where θ is the angle of inclination of the initial Velocity with horizontal component

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✭ According to the Question :

➞ Ke = ½mv²

➞ Ke = ½ × m × (u cosθ)²

m = Mass = 300 g = 300/1000 = 0.3 kg

u = Velocity = 25 m/s

θ = 45°

➞ Ke = ½ × 0.3 × (25 cos 45°)²

➞ Ke = ½ × 0.3 × 625 cos² 40°

➞ Ke = ½ × 0.3 × 625 × 0.586

➞ Ke = 54.937..

➞ Ke = 55 J

∴ The Kinetic Energy of the ball would be 55 J

Answer 2

$❥ᴀ᭄ɴsᴡᴇʀ$

Given

Mass = 300 g

Velocity = 25 m/s

θ = 40°

To Find

Kinetic Energy of the ball at the heighest point of trajectory

Solution

☯ Kinetic Energy = ½mv²

☯ v = u cosθ

Where θ is the angle of inclination of the initial Velocity with horizontal component

━━━━━━━━━━━━━━━━━━━━━━━━━

✭ According to the Question :

➞ Ke = ½mv²

➞ Ke = ½ × m × (u cosθ)²

m = Mass = 300 g = 300/1000 = 0.3 kg

u = Velocity = 25 m/s

θ = 45°

➞ Ke = ½ × 0.3 × (25 cos 45°)²

➞ Ke = ½ × 0.3 × 625 cos² 40°

➞ Ke = ½ × 0.3 × 625 × 0.586

➞ Ke = 54.937..

➞ Ke = 55 J

∴ The Kinetic Energy of the ball would be 55 J