Question

29. A body is projected at an angle 60° from
surface of earth with velocity 2gR. Final
K.E. of object is (symbols have usual
meanings)
(1) Zero
() mgR
(3) mgR
2
(4) Infinite​

Answer 1

Answer:

m g^2 R^2

Explanation:

K.E= 1 /2 mv^2

[v=2gR].

K.E= 1 / 2 m {( 2gR)^2 }

K.E= (mg^2)(R^2)

K.E= m g^2 R^2.

According to me,this should be the correct answer...