I. Tom throws a rock horizontally off of a 40 ,0 m high cliff.How fast did he throw the rock if it hits the ground 45.0 m from the base of the cliff?

Here's how to solve this projectile motion problem:**1. Analyze the Vertical Motion:*** **Known:** * Vertical displacement (Δy) = -40.0 m (negative because it's downwards) * Initial vertical velocity (Vy₀) = 0 m/s (since the rock is thrown horizontally) * Acceleration due to gravity (g) = -9.8 m/s² (negative because it's downwards)* **Unknown:** Time of flight (t)* **Equation:** Δy = Vy₀*t + (1/2)*g*t²* **Solve for t:** -40.0 m = (0 m/s)*t + (1/2)*(-9.8 m/s²)*t² -40.0 m = -4.9 m/s² * t² t² = (-40.0 m) / (-4.9 m/s²) t² = 8.16 s² t = √(8.16 s²) t ≈ 2.86 s**2. Analyze the Horizontal Motion:*** **Known:** * Horizontal displacement (Δx) = 45.0 m * Time of flight (t) = 2.86 s (calculated from vertical motion)* **Unknown:** Initial horizontal velocity (Vx₀)* **Equation:** Δx = Vx₀*t* **Solve for Vx₀:** 45.0 m = Vx₀ * 2.86 s Vx₀ = (45.0 m) / (2.86 s) Vx₀ ≈ 15.7 m/s**Answer:** Tom threw the rock horizontally with a speed of approximately 15.7 m/s.