Question

a projectile is fired with kinetic energy 1 KJ. If the range is maximum, what is its Kinetic energy, at the highest point?

Answer 1

If there is any calculation mistake please comment.

Answer 2

Answer:

kinetic energy=1000j

1/2mv^2 =1000j

v^2= 2000j/m

given that tetha=45° is maximum

therefore at highest point velocity=ucostetha

u/√2

ke. = 1/2m×u^2/2

=1/2 ×m × 2000/2m

= 500 j..........

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