Question

A projectile having an initial kinetic energy of 60 J is projected at an angle of 45° with the horizontal. What is the kinetic energy of the projectile at the highest of its trajectory?
1. 30 J
2. 15 J
3. 45 J
4. 60 J

Answer 1

Correct Answer - Option 1 : 30 J

The correct answer is 30 J.

• A particle is projected making an angle of 45^o with horizontal having kinetic energy K.
  • The kinetic energy at the highest point will be:
    • Kinetic Energy, \( K= mv^2/2\)​
  • At the highest point, a vertical component of velocity becomes zero and only the horizontal component is left that is given by:
    • Vcosθ
  • Let say Velocity = V
    • angle = 45^o
  • Horizontal Velocity = Vcos45^o
    • Cos45^o= 1/√2
  • ​Horizontal Velocity=  V/√2
  • Kinetic Energy = (1/2)mv²  = 60 Joule — (I) (Given)
  • V is V/√2
  • V²= (v/√2)²⇒ v²/2   —(II)
  • If we put V²/2 in equation (I) then,
    • Kinetic energy = (1/2)mv²/2 
  • As per equation (I),
    • (1/2)mv2  = 60 Joule
  • So, the Kinetic energy of projectile will be:
    • 60 Joule/2= 30 Joule.