A particle is projected at 60° to the horizontal. If its initial kinetic energy is KE₀ and the kinetic energy at maximum height is KE₀/x, find x.

Question

G1 Physics · Accepted Answer

At maximum height only the horizontal velocity component remains: v x = v v_x=v v x ​ = v cos60°=v/2. Hence K E t o p = ½ m ( v / 2 ) 2 = ( 1 / 4 ) ( ½ m v 2 ) = K E 0 / 4. E_top=½m(v/2)²=(1/4)(½mv²)=KE₀/4. E t ​ o p = ½ m ( v /2 ) 2 = ( 1/4 ) ( ½ m v 2 ) = K E 0 ​ /4. Therefore x=4.