if h be the maximum height of a projectile moving under the gravitational field of earth , then prove that its velocity of projection will be root2gh / sin theta where theta is angle of projection

Question

Answer

Proof of the Velocity of Projection of a Projectile

Maximum Height of a Projectile

Let's first understand the maximum height of a projectile moving under the gravitational field of earth:

The maximum height of a projectile is achieved at the highest point of its trajectory.

At this point, the vertical component of velocity becomes zero.

Using the formula for displacement in the vertical direction, we can find the maximum height:

Maximum height, h = (u^2 * sin^2(theta))/2g

Velocity of Projection of a Projectile

Now, let's prove that the velocity of projection of a projectile is root2gh / sin theta:

Using the formula for time of flight, we can express the time taken by a projectile to reach its maximum height:

Time of flight, t = 2u * sin(theta) / g

Using the formula for displacement in the horizontal direction, we can find the range of the projectile:

Range, R = u^2 * sin(2*theta) / g

Using the formula for average velocity, we can express the average velocity of the projectile as:

Average velocity, Vavg = R / t = u * sin(2*theta) / 2

Now, using the formula for the vertical component of velocity at the highest point:

Vertical component of velocity at highest point, V = u * sin(theta) - gt

Since the vertical component of velocity becomes zero at the highest point, we can equate V to zero and solve for u:

u = {gt}/{sin(theta)}

Substituting this value of u in the formula for average velocity, we get:

Vavg = {g * sin(2*theta)}/{2 * sin(theta)}

Using the formula for the maximum height of a projectile, we can express g in terms of h:

g = 2h / (sin(theta))^2

Substituting this value of g in the formula for average velocity, we get:

Vavg = {2h * sin(2*theta)}/{(sin(theta))^3}

Now, using the formula for the horizontal component of velocity, we can express u in terms of Vavg:

u = Vavg * 2 / sin(2*theta)

Substituting this value of u in the formula for the maximum height of a projectile, we get:

h = (Vavg^2 * sin^2(theta))/2g = (Vavg^2 * sin^2(theta))/4h

Answer

(vsinθ)² = (usinθ)² -2gh

But v=0

=> u²sin²θ = 2gh

=> u = √2gh/sinθ

Similar Class 11 Doubts

A particle of mass m is projected with a velocity v making an angle of 45 with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is --- A. Zero B. mv³/4√2g C. mv³/√2g D. m√(8gh³)?

When a projectile is projected at an angle with horizontal,find the angle of projection for its maximum horizontal range and the corresponding height achieved by the projectile?

Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with speed u at an angle 'theta' with the horizontal?

can a projectile be successful if angle if projection is less than 45°

A projectile thrown with velocity vo at an angle alpha to the horizontal has a range R. It will strike a vertical wall at a distance R/2 from point of projection with speed?

Top Courses for Class 11View all

Chemistry Class 11

CUET Mock Test: Humanities Subjects 2025

English Class 11

Mathematics (Maths) Class 11

Psychology Practice Tests: CUET Preparation

Top Courses for Class 11

Top Courses for Class 11

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams