**Projectile Motion Simulation**A ball is launched upward at an angle. The diagram below shows approximate values of the horizontal (\(v_x\)) and vertical (\(v_y\)) velocity vectors at a time of 0.0 seconds. Match the three timers to three of the provided velocity vectors by dragging a vector to a timer and releasing it. Assume gravity (\(g\)) is -10 m/s\(^2\).**Velocity Vectors and Timers:**- **Top Right Diagram:** - \(v_x = 20 \text{ m/s}\) - \(v_y = -20 \text{ m/s}\)Three timers are displayed to the right of the diagrams:1. **Time: 1.0 seconds**2. **Time: 2.0 seconds**3. **Time: 5.0 seconds****Six Velocity Vector Options:**1. \(v_x = 20 \text{ m/s}, v_y = 10 \text{ m/s}\)2. \(v_x = 20 \text{ m/s}, v_y = -30 \text{ m/s}\)3. \(v_x = 20 \text{ m/s}, v_y = -20 \text{ m/s}\)4. \(v_x = 10 \text{ m/s}, v_y = -20 \text{ m/s}\)5. \(v_x = 20 \text{ m/s}, v_y = 20 \text{ m/s}\)6. \(v_x = 0 \text{ m/s}, v_y = 20 \text{ m/s}\)To solve the simulation, determine which velocity vectors match each time interval by considering the effect of gravity on the vertical component over time.**Interactive Feature:**- **Check Answer**: Once you have matched the timers with the correct velocity vectors, click this button to verify your answers.

Name: **Projectile Motion Simulation**A ball is launched upward at an angle
Uploaded: 2024-03-20T06:57:47.000Z
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Description: **Projectile Motion Simulation**A ball is launched upward at an angle. The diagram below shows approximate values of the horizontal (\(v_x\)) and vertical (\(v_y\)) velocity vectors at a time of 0.0 seconds. Match the three timers to three of the pr

According to Newton's law of motion , the final velocity of the ball along the vertical axis at a time t is calculated by using the formula v=u+at , where u is the initial velocity , a is the acceleration and t is the time taken. A projectile motion , is the motion under gone by an object thrown at an angle with respect to the horizontal plane , and moving under the force of gravity alone. As the acceleration acting on the ball is the acceleration due to gravity alone , the velocity change will occur only in the vertical axis , thus the horizontal component of velocity will be a constant . Calculate the initial velocity of the ball by using the formula v=u+gt ,where u is the initial velocity , v=-20 m/s is the final velocity , t=5 s is time period and g=-10 m/s2 is the acceleration due to gravity. v=u+atu=v-at=-20 m/s--10 m/s2×5 s=-20 m/s+50 m/s=30 m/sCalculate the final velocity of the ball at time t=1 s by using the formula vy=u+gt ,where u=30 m/s is the initial velocity and g=-10 m/s2 is the acceleration due to gravity. vy=u+gt=30 m/s +-10 m/s2×1 s=30 m/s -10 m/s=20 m/sCalculate the final velocity of the ball at time t=2 s by using the formula vy=u+gt ,where u=30 m/s is the initial velocity and g=-10 m/s2 is the acceleration due to gravity. vy=u+gt=30 m/s +-10 m/s2×2 s=30 m/s -20 m/s=10 m/sCalculate the final velocity of the ball at time t=6 s by using the formula vy=u+gt ,where u=30 m/s is the initial velocity and g=-10 m/s2 is the acceleration due to gravity. vy=u+gt=30 m/s +-10 m/s2×6 s=30 m/s -60 m/s=-30 m/sThus the vertical and horizontal components of velocity at different time periods in the projectile motion is given below: At t=1 s, vx=20 m/svy=20 m/s . At t=2 s, vx=20 m/svy=10 m/s. At t=6 s, vx=20 m/svy=-30 m/s.

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A ball is launched upward at an angle. The diagram at the right shows approximate values of the horizontal and vertical velocity vectors and values at a time of 0.0 seconds. Time 5.0 Match the three timers to three of the Vx = 20 m/s provided velocity vectors. Drag to a vector and release. Assume g = -10 m/s/s. Vy = -20 m/s Time Vy = 10 m/s 1.0 Vx = 20 m/s Vx = 20 m/s Vx = 10 m/s Vy = -20 m/s Vy = -30 m/s Time 2.0 Vy = 20 m/s Vy = 20 m/s Vy = 0 m/s Vx = 0 m/s Vx = 20 m/s Vx = 20 m/s Time 6.0

A ball is launched upward at an angle. The diagram at the right shows approximate values of the horizontal and vertical velocity vectors and values at a time of 0.0 seconds. Time 5.0 Match the three timers to three of the Vx = 20 m/s provided velocity vectors. Drag to a vector and release. Assume g = -10 m/s/s. Vy = -20 m/s Time Vy = 10 m/s 1.0 Vx = 20 m/s Vx = 20 m/s Vx = 10 m/s Vy = -20 m/s Vy = -30 m/s Time 2.0 Vy = 20 m/s Vy = 20 m/s Vy = 0 m/s Vx = 0 m/s Vx = 20 m/s Vx = 20 m/s Time 6.0

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Projectile Motion

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**Projectile Motion Simulation**

A ball is launched upward at an angle. The diagram below shows approximate values of the horizontal (\(v_x\)) and vertical (\(v_y\)) velocity vectors at a time of 0.0 seconds. Match the three timers to three of the provided velocity vectors by dragging a vector to a timer and releasing it. Assume gravity (\(g\)) is -10 m/s\(^2\).

**Velocity Vectors and Timers:**

- **Top Right Diagram:**
- \(v_x = 20 \text{ m/s}\)
- \(v_y = -20 \text{ m/s}\)

Three timers are displayed to the right of the diagrams:

1. **Time: 1.0 seconds**
2. **Time: 2.0 seconds**
3. **Time: 5.0 seconds**

**Six Velocity Vector Options:**

1. \(v_x = 20 \text{ m/s}, v_y = 10 \text{ m/s}\)
2. \(v_x = 20 \text{ m/s}, v_y = -30 \text{ m/s}\)
3. \(v_x = 20 \text{ m/s}, v_y = -20 \text{ m/s}\)
4. \(v_x = 10 \text{ m/s}, v_y = -20 \text{ m/s}\)
5. \(v_x = 20 \text{ m/s}, v_y = 20 \text{ m/s}\)
6. \(v_x = 0 \text{ m/s}, v_y = 20 \text{ m/s}\)

To solve the simulation, determine which velocity vectors match each time interval by considering the effect of gravity on the vertical component over time.

**Interactive Feature:**

- **Check Answer**: Once you have matched the timers with the correct velocity vectors, click this button to verify your answers.

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Step 3 :Calculation of final velocity at t=1 s

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Step 4 :Calculation of final velocity at t=2 s

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Step 5 :Calculation of final velocity at t=6 s

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