Understanding Horizontal Projectile Motion
Projectile motion is a form of motion in which an object or particle is thrown near the Earth's surface and moves along a curved path under the action of gravity only. When an object is kicked horizontally from a cliff, we can decompose the motion into two independent components: horizontal and vertical.
Key Concepts
- Horizontal Motion: The velocity remains constant because there is no acceleration in the horizontal direction (ignoring air resistance). We use $x = v_x \cdot t$.
- Vertical Motion: The object starts with an initial vertical velocity of zero ($u_y = 0$) and undergoes constant acceleration due to gravity ($g \approx 9.8 \, m/s^2$). We use the kinematic equation $h = \frac{1}{2}gt^2$.
Solving the Problem
Given Data:
- Initial horizontal velocity ($v_x$) = $9 \, m/s$
- Height of the cliff ($h$) = $200 \, m$
- Acceleration due to gravity ($g$) = $9.8 \, m/s^2$ (assuming standard value)
Part i: Time taken to reach the ground
Using the vertical motion formula: $h = u_y t + \frac{1}{2}gt^2$ Since the stone is kicked horizontally, $u_y = 0$. Therefore: $200 = 0 + \frac{1}{2}(9.8)t^2$ $200 = 4.9t^2$ $t^2 = \frac{200}{4.9} \approx 40.82$ $t = \sqrt{40.82} \approx 6.39 \, s$
Part ii: Distance from the cliff (Range)
Using the horizontal motion formula: $R = v_x \cdot t$ $R = 9 \, m/s \cdot 6.39 \, s$ $R \approx 57.51 \, m$
Conclusion: The stone hits the ground after approximately $6.39$ seconds at a distance of $57.51$ meters from the base of the cliff.