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

Explanation:

A ball of mass 200g is thrown with a speed 20ms

−1 . The ball strikes a bat and rebounds along the same line at a speed 40ms −1 Variation of the interaction force, as long as the ball remains in contact with the bat, is as shown in Fig. Average force exerted by the bat on the ball is?

ANSWER

Given,

Mass, m=0.2kg

Final velocity, v f =40ms −1

Initial velocity, v i =−20ms −1

Change in momentum, ΔP=m(v f −vi)=0.2[40−(−20)]=12kgms −1

Average force, Fav = ΔTΔP

= 6×10 −312 =2000N

Hence, average force is 2000N

Answer 2

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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