Physics Problem: Maximum Height Calculation

This problem involves calculating the maximum height (\(H\)) reached by balls thrown vertically upwards under gravity (\(g\)). We need to use concepts from kinematics.

Analyzing the Conditions

• The second ball is thrown exactly when the first ball reaches its maximum height.
• The time difference between the throws of the two balls is given as 1 second.
• Acceleration due to gravity, \(g = 10 \text{ m/s}^2\).

Combining these two conditions, we infer that the time taken for the first ball to reach its maximum height is exactly 1 second. Let this time be \(t_{max}\).

So, \(t_{max} = 1 \text{ s}\).

Calculating Initial Velocity

For an object thrown vertically upwards, the velocity (\(v\)) at time (\(t\)) is given by \(v = u - gt\), where \(u\) is the initial velocity.

At the maximum height, the final velocity \(v = 0\). Therefore:

\(0 = u - g t_{max}\)

\(u = g t_{max}\)

Substituting the values:

\(u = (10 \text{ m/s}^2) \times (1 \text{ s})\)

\(u = 10 \text{ m/s}\)

This is the initial velocity with which the balls are thrown.

Calculating Maximum Height

The maximum height (\(H\)) reached by a projectile can be calculated using the formula:

\(H = \frac{u^2}{2g}\)

Substituting the calculated initial velocity (\(u = 10 \text{ m/s}\)) and the given value of \(g\):

\(H = \frac{(10 \text{ m/s})^2}{2 \times (10 \text{ m/s}^2)}\)

\(H = \frac{100 \text{ m}^2/\text{s}^2}{20 \text{ m/s}^2}\)

\(H = 5 \text{ m}\)

Since both balls are thrown under the same conditions (implied), they will attain the same maximum height.

Conclusion

The maximum height attained by each ball is 5 m.