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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?

Ask by Bird Schwartz. in South Africa
Feb 04,2025

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**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

Solución

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!

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Did you know that the concepts of motion and gravity were notably studied by Galileo Galilei in the Late Renaissance? He famously dropped two spheres of different weights from the Leaning Tower of Pisa, demonstrating that their time of descent was independent of their mass, paving the way for classical mechanics. This laid down the groundwork that allows us to solve these fascinating physics problems today! In real-world applications, the principles of motion and acceleration under gravity are crucial in numerous fields. For example, engineers utilize these calculations in designing safe buildings and bridges, while aerospace engineers need to understand flight dynamics for safely launching and landing spacecraft. Knowing how objects fall helps in everything from predicting the trajectory of sports balls to designing protective measures for structures and vehicles!

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