Projectile Motion Simulation
Stone thrown horizontally from a cliff
Theory
A hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 m/s. Neglecting air resistance, we can analyze the motion using projectile motion equations.
Given:
Initial height (y₀) = 490 m
Initial horizontal velocity (vₓ) = 15 m/s
Initial vertical velocity (v_y) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Equations of motion:
Horizontal position: \( x(t) = v_{ox} \cdot t \)
Vertical position: \( y(t) = y_0 + \frac{1}{2} a_y t^2 \)
Vertical velocity: \( v_y = -g \cdot t \)
Final speed: \( v = \sqrt{v_x^2 + v_y^2} \)
Calculations:
Time to reach ground: \( t = \sqrt{\frac{2 \cdot 490}{9.8}} = 10 \) s
Final vertical velocity: \( v_y = -9.8 \times 10 = -98 \) m/s
Final speed: \( \sqrt{15^2 + 98^2} = 99 \) m/s
