3. You throw a ball vertically upward from a height of 2 meters with an initial velocity of 10 meters per second. Use the idea that acceleration due to gravity (neglecting air resistance) is -32 feet/sec?, which is the same as -9.8 meters/sec2. a. What is the maximum height the ball will reach? Round your answer to the nearest meter, if necessary.

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### Physics Problem Set: Motion Under Gravity

#### Problem 3:
You throw a ball vertically upward from a height of 2 meters with an initial velocity of 10 meters per second. Use the idea that acceleration due to gravity (neglecting air resistance) is -32 feet/sec², which is the same as -9.8 meters/sec².

a. **Question**: What is the maximum height the ball will reach? Round your answer to the nearest meter, if necessary.

b. **Question**: How many seconds will go by before the ball hits the ground after being thrown? Round your answer to the nearest tenth of a second, if necessary.

**Detailed Explanation**: 
This problem involves the kinematic equations of motion under uniform acceleration due to gravity. The scenario described requires calculating the maximum height reached by the ball and the total time of flight until the ball hits the ground.

1. **Maximum Height Calculation**:
   - The initial height from which the ball is thrown is 2 meters.
   - The initial velocity (u) is 10 m/s.
   - The acceleration due to gravity (g) is -9.8 m/s².
   - The final velocity (v) at the maximum height is 0 m/s.

   Using the kinematic equation:
   \[
   v^2 = u^2 + 2as
   \]
   where \(s\) is the height gained.
   \[
   0 = (10)^2 + 2(-9.8)s
   \]
   \[
   0 = 100 - 19.6s
   \]
   \[
   19.6s = 100
   \]
   \[
   s = \frac{100}{19.6} \approx 5.1 \text{ meters}
   \]
   Thus, the maximum height \(H_{max}\) is the initial height plus \(s\):
   \[
   H_{max} = 2 + 5.1 \approx 7 \text{ meters}
   \]

2. **Total Time of Flight**:
   First, we need to determine the time taken to reach the maximum height:
   \[
   v = u + at
   \]
   \[
   0 = 10 + (-9.8)t
   \]
   \[
   9.8t = 10
Transcribed Image Text:### Physics Problem Set: Motion Under Gravity #### Problem 3: You throw a ball vertically upward from a height of 2 meters with an initial velocity of 10 meters per second. Use the idea that acceleration due to gravity (neglecting air resistance) is -32 feet/sec², which is the same as -9.8 meters/sec². a. **Question**: What is the maximum height the ball will reach? Round your answer to the nearest meter, if necessary. b. **Question**: How many seconds will go by before the ball hits the ground after being thrown? Round your answer to the nearest tenth of a second, if necessary. **Detailed Explanation**: This problem involves the kinematic equations of motion under uniform acceleration due to gravity. The scenario described requires calculating the maximum height reached by the ball and the total time of flight until the ball hits the ground. 1. **Maximum Height Calculation**: - The initial height from which the ball is thrown is 2 meters. - The initial velocity (u) is 10 m/s. - The acceleration due to gravity (g) is -9.8 m/s². - The final velocity (v) at the maximum height is 0 m/s. Using the kinematic equation: \[ v^2 = u^2 + 2as \] where \(s\) is the height gained. \[ 0 = (10)^2 + 2(-9.8)s \] \[ 0 = 100 - 19.6s \] \[ 19.6s = 100 \] \[ s = \frac{100}{19.6} \approx 5.1 \text{ meters} \] Thus, the maximum height \(H_{max}\) is the initial height plus \(s\): \[ H_{max} = 2 + 5.1 \approx 7 \text{ meters} \] 2. **Total Time of Flight**: First, we need to determine the time taken to reach the maximum height: \[ v = u + at \] \[ 0 = 10 + (-9.8)t \] \[ 9.8t = 10
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