Question

1. how high does the ball in #9 above travel?
2. justin, the famous race - car driver, is travelling at an initial velocity of 8.00 m/s and then accelerates uniformly at 10.0 m/s² for 5.00 seconds. how far does he travel in this time interval?
3. what is his final velocity in #11 above?
4. alexis throws a metal ball downward with a velocity of 20.0 m/s from a tall building. if the height of the building is 500 m, with what velocity does the ball hit the ground?
5. an electron in a vacuum tube is decelerated by a charged plate at a rate of - 7.00×10⁹ m/s². it decelerated during an interval of 6.00×10⁻³ seconds. the speed of the electron after the deceleration takes place is 2.00×10⁵ m/s. what is the speed of the electron before the deceleration? (this is an easy problem. list the givens and solve it!)
6. alexis is riding on a hot - air balloon and is ascending at a rate of 24.0 m/s. he accidentally knocks his bologna sandwich out of the balloon while at an altitude of 200 m. calculate the total time the bologna sandwich was in the air. neglect air resistance.

bonus.
vickia drops a heavy rock from a cliff 40 m above the water. it hits the water with a certain velocity and continues to sink to the bottom of the lake at this constant velocity. it reaches the bottom of the lake 12 seconds after it was dropped. how deep is the lake?

Explanation:

- ## Step1: Identify the kinematic equation
- \(x = v_0t+\frac{1}{2}at^{2}\)
- ## Step2: Substitute the values
- \(x=(8.00\times5.00)+\frac{1}{2}\times10.0\times(5.00)^{2}\)
- ## Step3: Calculate each term
- First term: \(8.00\times5.00 = 40\). Second - term: \(\frac{1}{2}\times10.0\times25 = 125\).
- ## Step4: Sum the terms
- \(x=40 + 125=165\)
2. **Answer for question 12**:
- We use the first - order kinematic equation \(v = v_0+at\), where \(v_0 = 8.00\ m/s\), \(a = 10.0\ m/s^{2}\), and \(t = 5.00\ s\).
- Substitute the values: \(v=8.00\ m/s+(10.0\ m/s^{2})\times(5.00\ s)\).
- \(v = 8.00\ m/s+50.0\ m/s=58.0\ m/s\).
- # Answer:
- \(58.0\ m/s\)
- # Explanation:
- ## Step1: Identify the kinematic equation
- \(v = v_0+at\)
- ## Step2: Substitute the values
- \(v = 8.00+10.0\times5.00\)
- ## Step3: Calculate the result
- \(v=8.00 + 50.0=58.0\)
3. **Answer for question 13**:
- We use the kinematic equation \(v^{2}=v_0^{2}+2ah\), where \(v_0 = 20.0\ m/s\), \(a = 9.8\ m/s^{2}\), and \(h = 500\ m\).
- Substitute the values: \(v^{2}=(20.0\ m/s)^{2}+2\times(9.8\ m/s^{2})\times500\ m\).
- First term: \((20.0\ m/s)^{2}=400\ m^{2}/s^{2}\).
- Second term: \(2\times9.8\times500\ m^{2}/s^{2}=9800\ m^{2}/s^{2}\).
- \(v^{2}=400 + 9800=10200\ m^{2}/s^{2}\).
- \(v=\sqrt{10200}\ m/s\approx101\ m/s\).
- # Answer:
- \(101\ m/s\)
- # Explanation:
- ## Step1: Identify the kinematic equation
- \(v^{2}=v_0^{2}+2ah\)
- ## Step2: Substitute the values
- \(v^{2}=(20.0)^{2}+2\times9.8\times500\)
- ## Step3: Calculate each term
- First term: \((20.0)^{2}=400\). Second - term: \(2\times9.8\times500 = 9800\).
- ## Step4: Solve for \(v\)
- \(v=\sqrt{400 + 9800}=\sqrt{10200}\approx101\)
4. **Answer for question 14**:
- We use the first - order kinematic equation \(v = v_0+at\), where \(v_0 = 2.00\times10^{5}\ m/s\), \(a=-7.00\times10^{9}\ m/s^{2}\), and \(t = 6.00\times10^{-3}\ s\).
- Substitute the values: \(v=(2.00\times10^{5}\ m/s)+(-7.00\times10^{9}\ m/s^{2})\times(6.00\times10^{-3}\ s)\).
- Second term: \((-7.00\times10^{9}\ m/s^{2})\times(6.00\times10^{-3}\ s)=-4.2\times10^{7}\ m/s\).
- Since \(| - 4.2\times10^{7}\ m/s|\gg2.00\times10^{5}\ m/s\), the electron stops and reverses direction. The speed after deceleration is \(|v| = 4.2\times10^{7}-2.00\times10^{5}=4.18\times10^{7}\ m/s\).
- # Answer:
- \(4.18\times10^{7}\ m/s\)
- # Explanation:
- ## Step1: Identify the kinematic equation
- \(v = v_0+at\)
- ## Step2: Substitute the values
- \(v=(2.00\times10^{5})+(-7.00\times10^{9})\times(6.00\times10^{-3})\)
- ## Step3: Calculate the second term
- \((-7.00\times10^{9})\times(6.00\times10^{-3})=-4.2\times10^{7}\)
- ## Step4: Calculate the final speed
- \(v=-4.2\times10^{7}+2.00\times10^{5}=-4.18\times10^{7}\), speed is \(4.18\times10^{7}\)
5. **Answer for question 15**:
- We use the kinematic equation \(h = v_0t+\frac{1}{2}gt^{2}\), where \(v_0 = 0\ m/s\) (the sandwich is just dropped), \(h=- 200\ m\) (downward direction is negative), and \(g=-9.8\ m/s^{2}\) (downward direction).
- The equation becomes \(-200\ m=0\times t+\frac{1}{2}\times(-9.8\ m/s^{2})\times t^{2}\).
- Rearranging for \(t\), we get \(t^{2}=\frac{2\times200}{9.8}\ s^{2}\).
- \(t=\sqrt{\frac{400}{9.8}}\ s\approx6.4\ s\).
- # Answer:
- \(6.4\ s\)
- # Explanation:
- ## Step1: Identify the kinematic equation
- \(h = v_0t+\frac{1}{2}gt^{2}\)
- ## Step2: Substitute the values
- \(-200=0\times t+\frac{1}{2}\times(-9.8)t^{2}\)
- ## Step3: Rearrange for \(t^{2}\)
- \(t^{2}=\frac{2\times200}{9.8}\)
- ## Step4: Solve for \(t\)
- \(t=\sqrt{\frac{400}{9.8}}\approx6.4\)
6. **Answer for BONUS**:
- First, we find the velocity of the rock when it hits the water. Using \(v^{2}=v_0^{2}+2gh_1\), where \(v_0 = 0\ m/s\), \(h_1 = 40\ m\), and \(g = 9.8\ m/s^{2}\).
- \(v^{2}=2\times9.8\times40\ m^{2}/s^{2}\), \(v=\sqrt{784}\ m/s = 28\ m/s\).
- Then, for the motion in the water, the rock moves at a constant velocity \(v = 28\ m/s\) for \(t = 12\ s\).
- The depth of the lake \(d=v\times t\), so \(d = 28\ m/s\times12\ s=336\ m\).
- # Answer:
- \(336\ m\)
- # Explanation:
- ## Step1: Find the velocity at water - surface
- \(v^{2}=2gh_1\)
- ## Step2: Calculate \(v\)
- \(v=\sqrt{2\times9.8\times40}=\sqrt{784}=28\)
- ## Step3: Calculate the depth of the lake
- \(d = v\times t\), \(d = 28\times12 = 336\)

Answer:

1. Answer for question 11:

• First, we use the second - order kinematic equation \(x = v_0t+\frac{1}{2}at^{2}\), where \(v_0 = 8.00\ m/s\), \(a = 10.0\ m/s^{2}\), and \(t = 5.00\ s\).
• Substitute the values into the formula:
• \(x=(8.00\ m/s)\times(5.00\ s)+\frac{1}{2}\times(10.0\ m/s^{2})\times(5.00\ s)^{2}\).
• First term: \((8.00\ m/s)\times(5.00\ s)=40\ m\).
• Second term: \(\frac{1}{2}\times(10.0\ m/s^{2})\times(5.00\ s)^{2}=\frac{1}{2}\times10.0\times25\ m = 125\ m\).
• Then \(x = 40\ m+125\ m=165\ m\).
• # Answer:
• \(165\ m\)
• # Explanation:
• ## Step1: Identify the kinematic equation
• \(x = v_0t+\frac{1}{2}at^{2}\)
• ## Step2: Substitute the values
• \(x=(8.00\times5.00)+\frac{1}{2}\times10.0\times(5.00)^{2}\)
• ## Step3: Calculate each term
• First term: \(8.00\times5.00 = 40\). Second - term: \(\frac{1}{2}\times10.0\times25 = 125\).
• ## Step4: Sum the terms
• \(x=40 + 125=165\)

1. Answer for question 12:

• We use the first - order kinematic equation \(v = v_0+at\), where \(v_0 = 8.00\ m/s\), \(a = 10.0\ m/s^{2}\), and \(t = 5.00\ s\).
• Substitute the values: \(v=8.00\ m/s+(10.0\ m/s^{2})\times(5.00\ s)\).
• \(v = 8.00\ m/s+50.0\ m/s=58.0\ m/s\).
• # Answer:
• \(58.0\ m/s\)
• # Explanation:
• ## Step1: Identify the kinematic equation
• \(v = v_0+at\)
• ## Step2: Substitute the values
• \(v = 8.00+10.0\times5.00\)
• ## Step3: Calculate the result
• \(v=8.00 + 50.0=58.0\)

1. Answer for question 13:

• We use the kinematic equation \(v^{2}=v_0^{2}+2ah\), where \(v_0 = 20.0\ m/s\), \(a = 9.8\ m/s^{2}\), and \(h = 500\ m\).
• Substitute the values: \(v^{2}=(20.0\ m/s)^{2}+2\times(9.8\ m/s^{2})\times500\ m\).
• First term: \((20.0\ m/s)^{2}=400\ m^{2}/s^{2}\).
• Second term: \(2\times9.8\times500\ m^{2}/s^{2}=9800\ m^{2}/s^{2}\).
• \(v^{2}=400 + 9800=10200\ m^{2}/s^{2}\).
• \(v=\sqrt{10200}\ m/s\approx101\ m/s\).
• # Answer:
• \(101\ m/s\)
• # Explanation:
• ## Step1: Identify the kinematic equation
• \(v^{2}=v_0^{2}+2ah\)
• ## Step2: Substitute the values
• \(v^{2}=(20.0)^{2}+2\times9.8\times500\)
• ## Step3: Calculate each term
• First term: \((20.0)^{2}=400\). Second - term: \(2\times9.8\times500 = 9800\).
• ## Step4: Solve for \(v\)
• \(v=\sqrt{400 + 9800}=\sqrt{10200}\approx101\)

1. Answer for question 14:

• We use the first - order kinematic equation \(v = v_0+at\), where \(v_0 = 2.00\times10^{5}\ m/s\), \(a=-7.00\times10^{9}\ m/s^{2}\), and \(t = 6.00\times10^{-3}\ s\).
• Substitute the values: \(v=(2.00\times10^{5}\ m/s)+(-7.00\times10^{9}\ m/s^{2})\times(6.00\times10^{-3}\ s)\).
• Second term: \((-7.00\times10^{9}\ m/s^{2})\times(6.00\times10^{-3}\ s)=-4.2\times10^{7}\ m/s\).
• Since \(| - 4.2\times10^{7}\ m/s|\gg2.00\times10^{5}\ m/s\), the electron stops and reverses direction. The speed after deceleration is \(|v| = 4.2\times10^{7}-2.00\times10^{5}=4.18\times10^{7}\ m/s\).
• # Answer:
• \(4.18\times10^{7}\ m/s\)
• # Explanation:
• ## Step1: Identify the kinematic equation
• \(v = v_0+at\)
• ## Step2: Substitute the values
• \(v=(2.00\times10^{5})+(-7.00\times10^{9})\times(6.00\times10^{-3})\)
• ## Step3: Calculate the second term
• \((-7.00\times10^{9})\times(6.00\times10^{-3})=-4.2\times10^{7}\)
• ## Step4: Calculate the final speed
• \(v=-4.2\times10^{7}+2.00\times10^{5}=-4.18\times10^{7}\), speed is \(4.18\times10^{7}\)

1. Answer for question 15:

• We use the kinematic equation \(h = v_0t+\frac{1}{2}gt^{2}\), where \(v_0 = 0\ m/s\) (the sandwich is just dropped), \(h=- 200\ m\) (downward direction is negative), and \(g=-9.8\ m/s^{2}\) (downward direction).
• The equation becomes \(-200\ m=0\times t+\frac{1}{2}\times(-9.8\ m/s^{2})\times t^{2}\).
• Rearranging for \(t\), we get \(t^{2}=\frac{2\times200}{9.8}\ s^{2}\).
• \(t=\sqrt{\frac{400}{9.8}}\ s\approx6.4\ s\).
• # Answer:
• \(6.4\ s\)
• # Explanation:
• ## Step1: Identify the kinematic equation
• \(h = v_0t+\frac{1}{2}gt^{2}\)
• ## Step2: Substitute the values
• \(-200=0\times t+\frac{1}{2}\times(-9.8)t^{2}\)
• ## Step3: Rearrange for \(t^{2}\)
• \(t^{2}=\frac{2\times200}{9.8}\)
• ## Step4: Solve for \(t\)
• \(t=\sqrt{\frac{400}{9.8}}\approx6.4\)

1. Answer for BONUS:

• First, we find the velocity of the rock when it hits the water. Using \(v^{2}=v_0^{2}+2gh_1\), where \(v_0 = 0\ m/s\), \(h_1 = 40\ m\), and \(g = 9.8\ m/s^{2}\).
• \(v^{2}=2\times9.8\times40\ m^{2}/s^{2}\), \(v=\sqrt{784}\ m/s = 28\ m/s\).
• Then, for the motion in the water, the rock moves at a constant velocity \(v = 28\ m/s\) for \(t = 12\ s\).
• The depth of the lake \(d=v\times t\), so \(d = 28\ m/s\times12\ s=336\ m\).
• # Answer:
• \(336\ m\)
• # Explanation:
• ## Step1: Find the velocity at water - surface
• \(v^{2}=2gh_1\)
• ## Step2: Calculate \(v\)
• \(v=\sqrt{2\times9.8\times40}=\sqrt{784}=28\)
• ## Step3: Calculate the depth of the lake
• \(d = v\times t\), \(d = 28\times12 = 336\)