135k views
3 votes
A 5 kg ball is thrown upward with a speed of 10 m/s. Calculate the maximum height attained by it and also find the potential energy when it reaches the highest point.

1 Answer

5 votes

Final answer:

The maximum height attained by the 5 kg ball thrown upward at 10 m/s is approximately 5.1 meters, and the potential energy at the highest point is approximately 250.05 Joules.

Step-by-step explanation:

To determine the maximum height attained by a 5 kg ball thrown upward with a speed of 10 m/s, we need to use the principle of conservation of energy. Initially, the ball has only kinetic energy (KE), which can be calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. At the highest point, this kinetic energy is completely converted into potential energy (PE), which is given by PE = m * g * h, where g is acceleration due to gravity (9.81 m/s²) and h is height.

Setting the initial kinetic energy equal to the potential energy at the highest point and solving for h gives us:

0.5 * m * v^2 = m * g * h

Simplifying, we get:

0.5 * (5 kg) * (10 m/s)² = (5 kg) * (9.81 m/s²) * h

h = (0.5 * 10²) / 9.81 = 5 / 9.81 m

h ≈ 5.1 m.

To find the potential energy at the highest point, simply substitute the height into the PE formula:

PE = (5 kg) * (9.81 m/s²) * 5.1 m ≈ 250.05 J.

User Viktor Svensson
by
8.4k points