1.

A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is:

Answer»

Zero
K/4
K/2
K

Solution :Here ENERGY at the START `1/2mv^(2)=K`
K.E. at the TOP `=1/2mv^(2)cos^(2)theta=Kcos^(2)60^@=K/4`


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