Ball Thrown from Tower At t1 = 0 a 1.0 kg ball is thrown from the top of a tall

Question

Ball Thrown from Tower At t1 = 0 a 1.0 kg ball is thrown from the top of a tall tower
with velocity v?1 ? (18 m/s)i ? (24 m/s)j What is the change in the potential energy of the
ballEarth system between t1 = 0 and t2 = 6.0 s?

Ball Thrown from Tower At t1 = 0 a 1.0 kg ball is thrown from the top of a tall tower with velocity v?1 ? (18 m/s)½i ? (24 m/s)½j What is the change in the potential energy of the ball½Earth system between t1 = 0 and t2 = 6.0 s?

Explanation / Answer

Let us make the things in direct numerals The velocity with which ball is thrown = 18i + 24j = Sqroot of ( 18^2 + 24^2) => v = Sqroot of (324+576) => v = 30m/s Let h be the height of the tower By Newton's laws of motion, Distance, s = vt + 1/2at^2 => s = 30 x 6 + 4.9 x 6 x 6 => s = 356.4m Therefore initial potential energy = mgh = 9.8h Potential energy after 6s = mgh = 9.8(h - 356.4) Change in PE = 9.8(h-356.4) - 9.8h = 9.8 { h - 356.4 - h } => Change in PE = 3492.72 J If it is the percentage change in PE that your answer demands then i will edit the answer

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