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.
1 Answers
caawiye Admin
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.