A ball with a mass of 0.7 kg is thrown straight upward, flies up to its maximum height, and then falls back down. If the ball was initially thrown with a kinetic energy of 250.1 joules,
how much potential energy will the ball have at its maximum height? Round your answer to
the tenths place.

250.05 J

Explanation:

From the question given above and, the following data were obtained:

Mass (m) = 0.7 kg

Kinetic energy (KE) = 250.1 J

Potential energy (PE) =?

Next, we shall determine the velocity of the ball. This can be obtained as follow:

Mass (m) = 0.7 kg

Kinetic energy (KE) = 250.1 J

Velocity (v) =?

KE = ½mv²

250.1 = ½ × 0.7 × v²

250.1 = 0.35 × v²

Divide both side by 0.35

v² = 250.1 / 0.35

Take the square root of both side

v = √(250.1 / 0.35)

v = 26.73 m/s

Next, we shall determine the maximum height reached by the ball. This can be obtained as follow:

Initial velocity (u) = 26.73 m/s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =.?

v² = u² – 2gh (since the ball is going against gravity)

0² = 26.73² – (2 × 9.8 × h)

0 = 714.4929 – 19.6h

Collect like terms:

0 – 714.4929 = – 19.6h

– 714.4929 = – 19.6h

Divide both side by –19.6

h = –714.4929 / –19.6

h = 36.45 m

Finally, we shall determine the potential energy of the ball at the maximum height. This can be obtained as follow:

Mass (m) = 0.7 kg

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) = 36.45 m

Potential energy (PE) =?

PE = mgh

PE = 0.7 × 9.8 × 36.45

PE = 250.05 J

Thus, the potential energy at the maximum height is 250.05 J.

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no. dot product   =0 when vecs at right angles