AL physics ~ Projectile motion ~~

Let v be the initial speed of the projectile.

As the horizontal motion is uniform, the horizontal speed

= vcos60* = v/2

Initially, K.E. = 1/2 mv2 = E

At the highest point, the projectile has only the horizontal component of the speed, which is equal to v/2

K.E. at the highest point

= 1/2 m(v/2)2

= 1/4 (1/2 mv2)

= E/4

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