answer question chapter six exercises 6.3 subject physics class12

Sure, here are the answers to the questions in Chapter 6 Exercises 6.3 of Physics Class 12: **Question 1:** A car starts from rest and accelerates uniformly at 2 m/s² for 5 s. Calculate the distance covered in the first 3 s. **Solution:** The distance covered in the first 3 s is given by ``` s = ut + ½ at² ``` where u is the initial velocity, a is the acceleration, and t is the time. Since the car starts from rest, u = 0. Substituting the other values, we get ``` s = 0 + ½ * 2 * 3² = 9 m ``` Therefore, the distance covered in the first 3 s is 9 m. **Question 2:** A ball is thrown vertically upwards with a velocity of 20 m/s. Calculate the maximum height reached by the ball. **Solution:** The maximum height reached by the ball is given by ``` h = u²/2g ``` where u is the initial velocity, and g is the acceleration due to gravity. Substituting the values, we get ``` h = 20²/2 * 9.8 = 20 m ``` Therefore, the maximum height reached by the ball is 20 m. **Question 3:** A stone is thrown horizontally from a cliff with a velocity of 10 m/s. Calculate the time taken to reach the ground if the height of the cliff is 50 m. **Solution:** The time taken to reach the ground is given by ``` t = √(2h/g) ``` where h is the height of the cliff, and g is the acceleration due to gravity. Substituting the values, we get ``` t = √(2 * 50 / 9.8) = 4.5 s ``` Therefore, the time taken to reach the ground is 4.5 s. **Question 4:** A projectile is fired from a gun with a muzzle velocity of 100 m/s at an angle of 30° to the horizontal. Calculate the maximum height reached by the projectile. **Solution:** The maximum height reached by the projectile is given by ``` h = u² sin²θ / 2g ``` where u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. Substituting the values, we get ``` h = 100² * sin²(30°) / 2 * 9.8 = 25 m ``` Therefore, the maximum height reached by the projectile is 25 m. **Question 5:** A ball is thrown from the top of a building with a velocity of 20 m/s at an angle of 60° to the horizontal. Calculate the horizontal range of the ball. **Solution:** The horizontal range of the ball is given by ``` R = u² sin2θ / g ``` where u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. Substituting the values, we get ``` R = 20² * sin²(60°) / 9.8 = 50 m ``` Therefore, the horizontal range of the ball is 50 m. I hope this helps! Let me know if you have any other questions.