1. A stone is dropped from the top of a building and hits the ground travelling at Checkpoint 2 $$ 45 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$. Ignore the effects of air resistance. Calculate the height of the building. 2. A ball is thrown upward at an unknown initial velocity. It takes $$ 3,2 \mathrm{~s} $$ to reach its highest point. Ignore the effects of air resistance. Calculate the initial velocity of the ball Exercise 2.1 Ignore the effects of air resistance. 1. A bomb falls out of an aircraft. To break the sound barrier of $$ 340 \mathrm{~m} . \mathrm{s}^{-1} $$ : a) how far does it need to fall? b) how long will this take? 2. A stone is thrown vertically upward from ground level with a velocity of $$ 25 \mathrm{~m} . \mathrm{s}^{-1} $$. Calculate: a) the maximum height reached. b) the time taken to reach its maximum height. 3. A stone is dropped from a bridge and is seen to splash into the water 3 s later. Calculate: a) the height of the bridge. b) the velocity with which the stone strikes the water. 4. A brick falls off a scaffold at a height of 80 m above the ground. Calculate: a) the magnitude of its velocity after falling for 2 s . b) the magnitude of its velocity when it hits the ground. c) the time taken to fall to the ground. 5. A stone, dropped from the top of a lighthouse, strikes the rocks below at a speed of $$ 50 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$. Calculate the height of the lighthouse. 6. A stone is thrown vertically upward and reaches a height of 10 m . a) What was the initial velocity of the stone as it left the thrower's hand?

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UpStudy AI · Accepted Answer

Certainly! Let's tackle each of these physics problems one by one. We'll use the basic equations of kinematics, assuming constant acceleration due to gravity ($$ g = 9.8 \, \text{m/s}^2 $$) and neglecting air resistance unless specified otherwise.

---

## **Problem 1**

**A stone is dropped from the top of a building and hits the ground traveling at $$ 45 \, \text{m/s} $$. Calculate the height of the building.**

### **Solution:**

When a stone is dropped (initial velocity $$ u = 0 $$), the final velocity ($$ v $$) upon impact and the distance fallen ($$ h $$) are related by the following kinematic equation:

$$ v^2 = u^2 + 2gh $$

Solving for $$ h $$:

$$ h = \frac{v^2 - u^2}{2g} $$

Given:
- $$ v = 45 \, \text{m/s} $$
- $$ u = 0 \, \text{m/s} $$
- $$ g = 9.8 \, \text{m/s}^2 $$

Plugging in the values:

$$ h = \frac{45^2 - 0}{2 \times 9.8} = \frac{2025}{19.6} \approx 103.31 \, \text{meters} $$

### **Answer:**
The height of the building is approximately **103.31 meters**.

---

## **Problem 2**

**A ball is thrown upward with an unknown initial velocity. It takes $$ 3.2 \, \text{s} $$ to reach its highest point. Calculate the initial velocity of the ball.**

### **Solution:**

At the highest point, the final velocity ($$ v $$) is 0. Using the equation:

$$ v = u + at $$

Where:
- $$ v = 0 \, \text{m/s} $$ (at the highest point)
- $$ a = -g = -9.8 \, \text{m/s}^2 $$ (acceleration due to gravity is downward)
- $$ t = 3.2 \, \text{s} $$

Solving for $$ u $$:

$$ 0 = u - 9.8 \times 3.2 $$
$$ u = 9.8 \times 3.2 = 31.36 \, \text{m/s} $$

### **Answer:**
The initial velocity of the ball is **31.36 m/s** upward.

---

## **Exercise 2.1**

### **Problem 1**

**A bomb falls out of an aircraft. To break the sound barrier of $$ 340 \, \text{m/s} $$:**

#### **a) How far does it need to fall?**

#### **b) How long will this take?**

### **Solution:**

Assuming the bomb is dropped from rest (initial velocity $$ u = 0 $$) and accelerates under gravity until it reaches $$ v = 340 \, \text{m/s} $$.

#### **a) Distance Fallen ($$ h $$)**

Using the equation:

$$ v^2 = u^2 + 2gh $$

Solving for $$ h $$:

$$ h = \frac{v^2 - u^2}{2g} = \frac{340^2}{2 \times 9.8} = \frac{115600}{19.6} \approx 5908.16 \, \text{meters} $$

#### **b) Time Taken ($$ t $$)**

Using:

$$ v = u + gt $$

Solving for $$ t $$:

$$ t = \frac{v - u}{g} = \frac{340}{9.8} \approx 34.69 \, \text{seconds} $$

### **Answer:**

1. **a)** The bomb needs to fall approximately **5908.16 meters**.

2. **b)** It will take around **34.69 seconds** to reach the speed of sound.

---

### **Problem 2**

**A stone is thrown vertically upward from ground level with a velocity of $$ 25 \, \text{m/s} $$. Calculate:**

#### **a) The maximum height reached.**

#### **b) The time taken to reach its maximum height.**

### **Solution:**

Given:
- Initial velocity $$ u = 25 \, \text{m/s} $$
- Final velocity at maximum height $$ v = 0 \, \text{m/s} $$
- Acceleration $$ a = -g = -9.8 \, \text{m/s}^2 $$

#### **a) Maximum Height ($$ h $$)**

Using:

$$ v^2 = u^2 + 2ah $$

Solving for $$ h $$:

$$ h = \frac{v^2 - u^2}{2a} = \frac{0 - 25^2}{2 \times (-9.8)} = \frac{-625}{-19.6} \approx 31.887 \, \text{meters} $$

#### **b) Time to Reach Maximum Height ($$ t $$)**

Using:

$$ v = u + at $$

Solving for $$ t $$:

$$ t = \frac{v - u}{a} = \frac{0 - 25}{-9.8} \approx 2.551 \, \text{seconds} $$

### **Answer:**

2. **a)** The maximum height reached is approximately **31.89 meters**.

2. **b)** The time taken to reach this height is approximately **2.55 seconds**.

---

### **Problem 3**

**A stone is dropped from a bridge and is seen to splash into the water 3 s later. Calculate:**

#### **a) The height of the bridge.**

#### **b) The velocity with which the stone strikes the water.**

### **Solution:**

Given:
- Time ($$ t $$) = 3 s
- Initial velocity ($$ u $$) = 0 m/s (stone is dropped)
- Acceleration $$ a = g = 9.8 \, \text{m/s}^2 $$

#### **a) Height of the Bridge ($$ h $$)**

Using:

$$ h = ut + \frac{1}{2}gt^2 = 0 + \frac{1}{2} \times 9.8 \times 3^2 = 4.9 \times 9 = 44.1 \, \text{meters} $$

#### **b) Velocity upon Impact ($$ v $$)**

Using:

$$ v = u + gt = 0 + 9.8 \times 3 = 29.4 \, \text{m/s} $$

### **Answer:**

3. **a)** The height of the bridge is **44.1 meters**.

3. **b)** The velocity upon striking the water is **29.4 m/s** downward.

---

### **Problem 4**

**A brick falls off a scaffold at a height of 80 m above the ground. Calculate:**

#### **a) The magnitude of its velocity after falling for 2 s.**

#### **b) The magnitude of its velocity when it hits the ground.**

#### **c) The time taken to fall to the ground.**

### **Solution:**

Given:
- Height $$ h = 80 \, \text{m} $$
- Initial velocity $$ u = 0 \, \text{m/s} $$
- Acceleration $$ a = g = 9.8 \, \text{m/s}^2 $$

#### **a) Velocity After 2 Seconds ($$ v $$)**

Using:

$$ v = u + at = 0 + 9.8 \times 2 = 19.6 \, \text{m/s} $$

#### **b) Velocity Upon Impact**

First, find the velocity upon impact using:

$$ v^2 = u^2 + 2gh $$

$$ v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 80} = \sqrt{1568} \approx 39.6 \, \text{m/s} $$

#### **c) Time to Fall to the Ground ($$ t $$)**

Using:

$$ h = ut + \frac{1}{2}gt^2 $$

$$ 80 = 0 + \frac{1}{2} \times 9.8 \times t^2 $$

$$ t^2 = \frac{160}{9.8} \approx 16.3265 $$

$$ t \approx \sqrt{16.3265} \approx 4.04 \, \text{seconds} $$

### **Answer:**

4. **a)** After 2 seconds, the brick's velocity is **19.6 m/s** downward.

4. **b)** Upon hitting the ground, its velocity is approximately **39.6 m/s** downward.

4. **c)** It takes about **4.04 seconds** to fall to the ground.

---

### **Problem 5**

**A stone, dropped from the top of a lighthouse, strikes the rocks below at a speed of $$ 50 \, \text{m/s} $$. Calculate the height of the lighthouse.**

### **Solution:**

Given:
- Final velocity ($$ v $$) = 50 m/s
- Initial velocity ($$ u $$) = 0 m/s
- Acceleration $$ a = g = 9.8 \, \text{m/s}^2 $$

Using:

$$ v^2 = u^2 + 2gh $$

Solving for $$ h $$:

$$ h = \frac{v^2 - u^2}{2g} = \frac{50^2}{2 \times 9.8} = \frac{2500}{19.6} \approx 127.55 \, \text{meters} $$

### **Answer:**
The height of the lighthouse is approximately **127.55 meters**.

---

### **Problem 6**

**A stone is thrown vertically upward and reaches a height of 10 m. What was the initial velocity of the stone as it left the thrower's hand?**

### **Solution:**

Given:
- Maximum height ($$ h $$) = 10 m
- Final velocity at maximum height ($$ v $$) = 0 m/s
- Acceleration $$ a = -g = -9.8 \, \text{m/s}^2 $$

Using the equation:

$$ v^2 = u^2 + 2ah $$

$$ 0 = u^2 + 2 \times (-9.8) \times 10 $$

$$ u^2 = 2 \times 9.8 \times 10 = 196 $$

$$ u = \sqrt{196} = 14 \, \text{m/s} $$

### **Answer:**
The initial velocity of the stone was **14 m/s** upward.

---

## **Summary of Answers**

1. **Height of the building:** ~103.31 meters
2. **Initial velocity of the ball:** 31.36 m/s
3. 
   - a) Bomb falls ~5908.16 meters
   - b) Takes ~34.69 seconds
4. 
   - a) Max height 31.89 meters
   - b) Time to max height ~2.55 seconds
5. 
   - a) Bridge height 44.1 meters
   - b) Impact velocity 29.4 m/s
6. 
   - a) Velocity after 2s: 19.6 m/s
   - b) Impact velocity: 39.6 m/s
   - c) Time to ground: 4.04 seconds
7. **Height of lighthouse:** ~127.55 meters
8. **Initial velocity of stone:** 14 m/s upward

Feel free to ask if you need further explanations or assistance with these problems!
**Solutions:** 1. **Height of the building:** ~103.31 meters 2. **Initial velocity of the ball:** 31.36 m/s 3. - a) Bomb falls ~5908.16 meters - b) Takes ~34.69 seconds 4. - a) Max height 31.89 meters - b) Time to max height ~2.55 seconds 5. - a) Bridge height 44.1 meters - b) Impact velocity 29.4 m/s 6. - a) Velocity after 2s: 19.6 m/s - b) Impact velocity: 39.6 m/s - c) Time to ground: 4.04 seconds 7. **Height of lighthouse:** ~127.55 meters 8. **Initial velocity of stone:** 14 m/s upward

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