An object is launched at a velocity of 25 m/s in a direction making an angle of 20° upward with the horizontal.What is the maximum height reached by the object?What is the horizontal range (maximum x above ground) of the object? ### PHY 150 PROBLEM SET 3#### Projectile Motion Equations1. **Horizontal Velocity:** \[ v_x = v_o \cos \theta = \text{constant} \]2. **Vertical Velocity:** \[ v_y = v_o \sin \theta \]3. **Vertical Velocity at some time \( t \):** \[ v_y = v_o (\sin \theta) - gt \] where \( g = 9.8 \frac{m}{s^2} \)4. **Vertical Velocity at height \( y \):** \[ v_y^2 = (v_o \sin \theta)^2 - 2g \Delta y \]5. **Horizontal Distance Traveled:** \[ \Delta x = v_o (\cos \theta) t \]6. **Vertical Distance Traveled:** \[ \Delta y = (v_o \sin \theta)t - \frac{1}{2}gt^2 \]7. **Time to Highest Point:** \[ t = \frac{v_o \sin \theta}{g} \]8. **Total Flight Time:** \[ t = \frac{2v_o \sin \theta}{g} \]These equations are fundamental in understanding the principles of projectile motion. They cover various aspects of projectile trajectory including horizontal and vertical components of velocity, displacement, and time calculations. * **Horizontal Velocity ( \( v_x \) ):** This remains constant throughout the projectile’s flight.* **Vertical Velocity ( \( v_y \) ):** This changes due to the influence of gravity.* **Vertical Velocity at some time \( t \):** This is the initial vertical velocity minus the product of gravitational acceleration ( \( g \) ) and time ( \( t \) ).* **Vertical Velocity at height \( y \):** This relation uses the squared initial vertical velocity, gravitational acceleration, and the vertical distance \( \Delta y \).* **Horizontal Distance Traveled ( \( \Delta x \) ):** The distance covered horizontally over time.* **Vertical Distance Traveled ( \( \Delta y \) ):** The distance covered vertically over time, adjusted for gravitational effects.* **Time to Highest Point:** The time it takes for the projectile to reach its maximum height.* **Total Flight Time:** The total duration the projectile is in

Name: An object is launched at a velocity of 25 m/s in a direction making an
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Description: An object is launched at a velocity of 25 m/s in a direction making an angle of 20° upward with the horizontal.What is the maximum height reached by the object?What is the horizontal range (maximum x above ground) of the object? ### PHY 150 PROBLEM S