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.

50 inches should be the resultant vector

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong