Introduction to Horizontal Projectile Motion
Horizontal projectile motion is a classic physics problem where an object is given an initial horizontal velocity and allowed to fall under the influence of gravity. The key to solving these problems is the principle of independence of motion: the horizontal and vertical motions are independent of each other.
The Physics Principles
- Horizontal Motion: Since there is no air resistance, the horizontal velocity ($v_x$) remains constant throughout the flight: $v_x = u = 9.8 \text{ m/s}$.
- Vertical Motion: The object undergoes constant acceleration due to gravity ($g = 9.8 \text{ m/s}^2$). The vertical velocity ($v_y$) increases as the object falls, following $v_y = u_y + gt$. Since the projection is purely horizontal, the initial vertical velocity $u_y = 0$.
Step-by-Step Solution
Given data:
- Height of tower ($h$) = $100 \text{ m}$
- Initial horizontal velocity ($v_x$) = $9.8 \text{ m/s}$
- Initial vertical velocity ($u_y$) = $0 \text{ m/s}$
- Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$
Step 1: Find the vertical velocity upon impact ($v_y$)
We use the kinematic equation: $v_y^2 = u_y^2 + 2gh$
$v_y^2 = 0^2 + 2 \cdot 9.8 \cdot 100$ $v_y^2 = 1960$ $v_y = \sqrt{1960} \approx 44.27 \text{ m/s}$
Step 2: Calculate the resultant velocity ($v$)
The resultant velocity is the vector sum of horizontal and vertical components:
$v = \sqrt{v_x^2 + v_y^2}$ $v = \sqrt{(9.8)^2 + (44.27)^2}$ $v = \sqrt{96.04 + 1960}$ $v = \sqrt{2056.04} \approx 45.34 \text{ m/s}$
Conclusion
The object strikes the ground with a final velocity of approximately $45.34 \text{ m/s}$.