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

At highest position, occurs only horizontal component of velocity

e.g., V = ucosθ, here θ is the angle of inclination of initial velocity with horizontal

So, V = 20 cos40°

Know, kinetic energy = 1/2 mV²

= 1/2 × (200/1000) kg × {20cos40°}² J

= 1/2 × 0.2 × 400cos²40° J

= 0.1 × 400 × 0.586 [ ∵ cos²40° = 0.586 ]

= 40 × 0.586 J

= 23.54 J

Answer 2

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