for a projectile launched upwards from level ground with initial velocity u and angle of projection 30°with vertical, height of projectile above ground will be greater than horizontal distance covered upto time.

When we analyze the motion of a projectile launched at an angle, it's essential to break down its components: vertical and horizontal. In your scenario, the projectile is launched upwards from level ground with an initial velocity \( u \) at an angle of 30° with the vertical. This means the angle with the horizontal is 60° (since the total angle in a straight line is 90°). Let's delve into the details to understand why the height of the projectile can exceed the horizontal distance covered at a given time.

Breaking Down the Motion

The motion of the projectile can be described using two key components: the vertical motion and the horizontal motion. We can use trigonometric functions to find the initial vertical and horizontal velocities.

• Vertical Component (Vy): This is calculated as \( V_y = u \cdot \cos(30°) = u \cdot \frac{\sqrt{3}}{2} \).
• Horizontal Component (Vx): This is calculated as \( V_x = u \cdot \sin(30°) = u \cdot \frac{1}{2} \).

Vertical Motion Analysis

The height \( h \) of the projectile at any time \( t \) can be determined using the formula:

\( h = V_y \cdot t - \frac{1}{2} g t^2 \)

Here, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). Plugging in the vertical component:

\( h = \left( u \cdot \frac{\sqrt{3}}{2} \right) t - \frac{1}{2} g t^2 \)

Horizontal Motion Analysis

The horizontal distance \( d \) covered by the projectile at time \( t \) is given by:

\( d = V_x \cdot t \)

Substituting the horizontal component:

\( d = \left( u \cdot \frac{1}{2} \right) t \)

Comparing Height and Horizontal Distance

To determine when the height exceeds the horizontal distance, we need to compare the two equations derived above:

\( \left( u \cdot \frac{\sqrt{3}}{2} \right) t - \frac{1}{2} g t^2 > \left( u \cdot \frac{1}{2} \right) t \)

Rearranging this inequality gives:

\( \left( u \cdot \frac{\sqrt{3}}{2} - \frac{u}{2} \right) t > \frac{1}{2} g t^2 \)

Factoring out \( t \) (assuming \( t > 0 \)), we can simplify it to:

\( \left( u \cdot \left( \frac{\sqrt{3}}{2} - \frac{1}{2} \right) \right) > \frac{1}{2} g t \)

This shows that as long as the initial velocity \( u \) is sufficiently large, the height can indeed be greater than the horizontal distance covered at certain times during the projectile's flight. The projectile will reach a maximum height before descending, and during its ascent, the height can surpass the horizontal distance.

Conclusion

In summary, the relationship between height and horizontal distance for a projectile launched at an angle depends on the initial velocity and the effects of gravity. At certain points in its trajectory, particularly during the ascent, the height can exceed the horizontal distance covered. This dynamic interplay of vertical and horizontal motion is a fascinating aspect of projectile motion that illustrates the principles of physics in action.