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2+2=4, then divide it by 0

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A ball is thrown from a 50-meter tall building at 20 m/s at a 60° angle. Find when and where it hits the ground.

A ball is thrown from a 50-meter tall building at 20 m/s at a 60° angle. Find when and where it hits the ground.

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A ball is thrown from a 50-meter tall building at 20 m/s at a 60° angle. Find when and where it hits the ground.

A ball is thrown from a 50-meter tall building at 20 m/s at a 60° angle. Find when and where it hits the ground.

A ball is thrown from a 50-meter tall building at 20 m/s at a 60° angle. Find when and where it hits the ground.

**Problem Statement** A car moves horizontally with a constant velocity of 20 m/s. From this moving car, a ball is thrown straight upward with an initial vertical velocity of 10 m/s relative to the car. The ball experiences two forces: gravity acting downward and air resistance proportional to its velocity. The drag force always acts opposite to the direction of motion. The ball has a mass of 0.2 kg, the linear drag coefficient is 0.1 kg/s, and gravitational acceleration is 9.8 m/s². At time t = 0, the ball is at position x = 0, y = 0 with initial velocities v_x = 20 m/s and v_y = 10 m/s. You must model the ball’s horizontal and vertical position and velocity over time using the given physical forces and determine whether the ball ever returns to the moving car. The car continues moving at constant velocity, so its position is x_c(t) = 20t. Your task is to compute the motion of the ball, find the time at which it lands (when y returns to 0), and check whether its horizontal position matches the car’s position at that moment. Use the equations of motion under linear drag: dv_x/dt = -(k/m) v_x dv_y/dt = -g -(k/m) v_y dx/dt = v_x dy/dt = v_y

**Problem Statement** A car moves horizontally with a constant velocity of 20 m/s. From this moving car, a ball is thrown straight upward with an initial vertical velocity of 10 m/s relative to the car. The ball experiences two forces: gravity acting downward and air resistance proportional to its velocity. The drag force always acts opposite to the direction of motion. The ball has a mass of 0.2 kg, the linear drag coefficient is 0.1 kg/s, and gravitational acceleration is 9.8 m/s². At time t = 0, the ball is at position x = 0, y = 0 with initial velocities v_x = 20 m/s and v_y = 10 m/s. You must model the ball’s horizontal and vertical position and velocity over time using the given physical forces and determine whether the ball ever returns to the moving car. The car continues moving at constant velocity, so its position is x_c(t) = 20t. Your task is to compute the motion of the ball, find the time at which it lands (when y returns to 0), and check whether its horizontal position matches the car’s position at that moment. Use the equations of motion under linear drag: dv_x/dt = -(k/m) v_x dv_y/dt = -g -(k/m) v_y dx/dt = v_x dy/dt = v_y

Problem Statement A car moves horizontally with constant velocity. A ball is thrown vertically upward from the car. The ball is subjected to gravity and air resistance proportional to its velocity. Model the position and velocity of the ball over time and determine whether it returns to the car. --- Given Car velocity (constant): v_c = 20\ \text{m/s} Initial vertical velocity of the ball (relative to car): v_{y0} = 10\ \text{m/s} Gravitational acceleration: g = 9.8\ \text{m/s}^2 Ball mass: m = 0.2\ \text{kg} Linear drag coefficient: k = 0.1\ \text{kg/s} Initial conditions: x(0) = 0,\quad y(0) = 0 v_x(0) = 20,\quad v_y(0) = 10  --- Forces Acting on the Ball Gravity: \vec{F}_g = (0,\,-mg) Air resistance (linear drag): \vec{F}_d = -k\vec{v} --- Equations of Motion (Vector Form) m\frac{d\vec{v}}{dt} = -k\vec{v} + (0,-mg) --- Component Form Horizontal direction m\frac{dv_x}{dt} = -k v_x Vertical direction m\frac{dv_y}{dt} = -k v_y - mg --- Velocity Solutions Horizontal velocity v_x(t) = v_c\,e^{-(k/m)t} v_x(t) = 20\,e^{-0.5t} --- Vertical velocity v_y(t) = \left(v_{y0} + \frac{mg}{k}\right)e^{-(k/m)t} - \frac{mg}{k} v_y(t) = (10 + 19.6)e^{-0.5t} - 19.6 --- Position Equations Horizontal position x(t) = \int v_x(t)\,dt x(t) = \frac{m v_c}{k}\left(1 - e^{-(k/m)t}\right) x(t) = 40\left(1 - e^{-0.5t}\right) --- Vertical position y(t) = \int v_y(t)\,dt y(t) = \left(\frac{m}{k}\right)\left(v_{y0} + \frac{mg}{k}\right) \left(1 - e^{-(k/m)t}\right) - \frac{mg}{k}t y(t) = 59.2(1 - e^{-0.5t}) - 19.6t --- Car Motion x_c(t) = v_c t = 20t --- Key Simulation Result The ball does NOT return to the car. Why: Horizontal velocity decays exponentially Car continues at constant speed Relative horizontal displacement increases --- Condition to Check in Simulation Ball lands when: y(t_f) = 0 At that time: x_b(t_f) < x_c(t_f) ✔ Ball falls behind the car --- What Your Simulation Must Show 1. Exponentially decaying horizontal velocity 2. Asymmetric vertical motion (slower rise, faster fall) 3. Ball landing behind the car 4. Energy loss over time --- Numerical Integration Form (for code) Use this if you’re stepping frame-by-frame: \frac{dx}{dt} = v_x \frac{dy}{dt} = v_y \frac{dv_x}{dt} = -\frac{k}{m}v_x \frac{dv_y}{dt} = -g - \frac{k}{m}v_y ✔ Perfect for Euler / Verlet / RK4 --- Upgrade to Realistic Drag (Optional v2) Replace drag force with: \vec{F}_d = -c|\vec{v}|\vec{v} Then: \frac{dv_x}{dt} = -\frac{c}{m}|\vec{v}|v_x \frac{dv_y}{dt} = -g - \frac{c}{m}|\vec{v}|v_y (No closed-form solution → numerical only) --- Keywords for Your Prototype Linear drag model Non-conservative force Exponential velocity decay Coupled differential equations Numerical integration