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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
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if h be the maximum height of a projectile moving under the gravitatio...
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



Community Answer
if h be the maximum height of a projectile moving under the gravitatio...
(vsinθ)² = (usinθ)² -2gh

But v=0

=> u²sin²θ = 2gh

=> u = √2gh/sinθ
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