A PROJECTILE HAVING INITIAL KINETIC ENERGY 60J IS PROJECTED AT AN ANGLE OF 45DEGREE WITH THE HORIZONTAL WHAT IS THE KINETIC ENERGY OF THE PROJECTED AT THE HIGHEST POINT OF ITS TRAJECTORY?
Answers
Given : A PROJECTILE HAVING INITIAL KINETIC ENERGY 60J IS PROJECTED AT AN ANGLE OF 45DEGREE WITH THE HORIZONTAL
To Find : WHAT IS THE KINETIC ENERGY OF THE PROJECTED AT THE HIGHEST POINT OF ITS TRAJECTORY?
Solution:
Let say Velocity = V
angle = 45°
Horizontal Velocity = Vcos45° = V/√2
Vertical Velocity = Vsin45° = V√2
Kinetic Energy = (1/2)mV² = 60 Joule
Horizontal Component = 30 joule and Vertical component = 30 joule
( as Horizontal and vertical velocities are same)
At Highest point Vertical component becomes 0 as vertical velocity is zero
Hence only kinetic energy left is Horizontal component
Hence Kinetic energy at the highest point = 30 Joule
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