Question

A ball is thrown vertically upward. It has a speed of 10m/s when it has reached one half of its maximum height. How high does the ball rise?(Taking g=10 m/s²)

• A. 6m
• B. 10m
• C. 14m
• D. 18m

Answer 1

The Correct Option is B

To solve this problem, we need to determine the maximum height reached by the ball. We know the velocity of the ball is 10 m/s at half of its maximum height. Let's use the equations of motion under constant acceleration to find the solution.

Step 1: Understand the given data and apply the energy conservation concept.

We know:

• The initial velocity of the ball when it is thrown upward: u
• The final velocity of the ball at maximum height: v = 0 \, \text{m/s}
• Acceleration due to gravity: g = -10 \, \text{m/s}^{2}

We also have the velocity at half the maximum height:

• v = 10 \, \text{m/s}

Let H be the maximum height reached by the ball.

Step 2: Use the kinetic and potential energy relation.

At the maximum height, all the kinetic energy is converted into potential energy.

Using the energy conservation between the point where velocity is 10 m/s and the maximum height:

Initial kinetic energy at \frac{H}{2} and potential energy relation:

\frac{1}{2}mv^2 = mg \cdot \frac{H}{2}

Substitute the known values:

\frac{1}{2}m(10)^2 = m \cdot 10 \cdot \frac{H}{2}

Cancel the mass m and simplify:

50 = 5H

Therefore,

H = 10

meters.

Conclusion: The maximum height (H) reached by the ball is 10 meters. Thus, the correct answer is 10m.