Class 11 Exam  >  Class 11 Questions  >  A projectile fired with a velocity 'u' making... Start Learning for Free
A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.?
Most Upvoted Answer
A projectile fired with a velocity 'u' making an angle with the horizo...
Community Answer
A projectile fired with a velocity 'u' making an angle with the horizo...
Projectile Motion

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration due to gravity. The path followed by a projectile is called its trajectory.

Deriving the Expression for Maximum Height Attained

To derive the expression for the maximum height attained by a projectile, we need to consider the motion of the projectile in the vertical direction. The projectile is subject to the acceleration due to gravity (g) in the vertical direction. The initial velocity of the projectile can be resolved into horizontal and vertical components.

Let u be the initial velocity of the projectile, θ be the angle of projection with the horizontal, and h be the maximum height attained by the projectile.

Using the equations of motion, we have:

v = u + gt (where v is the final velocity)

At maximum height, v = 0, and t is the time taken to reach the maximum height.

So, 0 = u sin θ - gt

Solving for t, we get:

t = u sin θ / g

Now, using the equation of motion, we have:

h = ut sin θ - 1/2 gt^2

Substituting the value of t, we get:

h = u^2 sin^2 θ / 2g

Therefore, the expression for the maximum height attained by a projectile is:

h = u^2 sin^2 θ / 2g

Deriving the Expression for Horizontal Range

To derive the expression for the horizontal range, we need to consider the motion of the projectile in the horizontal direction. The projectile is subject to no acceleration in the horizontal direction, and it moves with uniform velocity.

Let R be the horizontal range of the projectile.

Using the equations of motion, we have:

R = ut cos θ

Substituting the value of t from above, we get:

R = u^2 sin 2θ / g

Therefore, the expression for the horizontal range of a projectile is:

R = u^2 sin 2θ / g

Deriving the Expression for Time of Flight

The time of flight is the time taken by the projectile to reach the ground. To derive the expression for the time of flight, we need to consider the motion of the projectile in the vertical direction.

Using the equations of motion, we have:

y = ut sin θ - 1/2 gt^2

where y is the displacement of the projectile in the vertical direction.

At the highest point, y = h, and t is the time taken to reach the highest point.

So, h = u^2 sin^2 θ / 2g

Solving for t, we get:

t = 2u sin θ / g

Therefore, the expression for the time of flight of a projectile is:

t = 2u sin θ / g

Conclusion

In conclusion, the expressions for the maximum height attained, horizontal range, and time of flight of a projectile are derived by analyzing the motion of the projectile in the vertical and horizontal directions. These expressions are useful in analyzing the motion of projectiles in various applications, such as ballistics and sports.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
Explore Courses for Class 11 exam

Top Courses for Class 11

A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.?
Question Description
A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.?.
Solutions for A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? in English & in Hindi are available as part of our courses for Class 11. Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? defined & explained in the simplest way possible. Besides giving the explanation of A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.?, a detailed solution for A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? has been provided alongside types of A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? theory, EduRev gives you an ample number of questions to practice A projectile fired with a velocity 'u' making an angle with the horizontal . Derive an expression for maximum height attained , horizontal range,time of flight.? tests, examples and also practice Class 11 tests.
Explore Courses for Class 11 exam

Top Courses for Class 11

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev