Understanding Maximum Height in Projectile Motion

When a projectile is fired, it follows a curved path called a trajectory. The highest point reached by the projectile during its flight is known as the maximum height.

The maximum height depends on the initial velocity and the angle at which the projectile is launched.

Given Information

• Initial speed (\(V_0\)) = \(V_0\) m/s
• Launch angle (\(\theta\)) = 30° from the horizontal
• Acceleration due to gravity = \(g\)

Formula for Maximum Height

The formula used to calculate the maximum height (\(H\)) attained by a projectile launched with initial velocity \(V_0\) at an angle \(\theta\) with the horizontal is:

\[H = \frac{{V_0}^2 \sin^2\theta}{2g}\]

Calculating the Maximum Height

We are given the launch angle \(\theta = 30°\). We need to find the value of \(\sin(30°)\) and then square it.

• We know that \(\sin(30°) = \frac{1}{2}\).
• So, \(\sin^2(30°) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).

Now, substitute the values of \(V_0\), \(\sin^2\theta\), and \(g\) into the maximum height formula:

\[H = \frac{{V_0}^2 \times \left(\frac{1}{4}\right)}{2g}\]

\[H = \frac{\frac{{V_0}^2}{4}}{2g}\]

To simplify the expression, we can write it as:

\[H = \frac{{V_0}^2}{4} \times \frac{1}{2g}\]

\[H = \frac{{V_0}^2}{8g}\]

Therefore, the maximum height attained by the projectile is \(\frac{{{V_0}^2}}{{8g}}\).

Conclusion

The maximum height reached by the projectile fired at an angle of 30° with a speed of \(V_0\) m/s is \(\frac{{{V_0}^2}}{{8g}}\).