Projectile Fun!

Classical Mechanics Level 2

Find the average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory, with the projection speed is u u and angle of projection with the horizontal θ \theta .

u 2 1 + cos ⁡ 2 θ \frac { u }{ 2 } \sqrt { 1+\cos ^{ 2 }{ \theta } } u 2 cos ⁡ θ \frac { u }{ 2 } \cos { \theta } u 2 2 + cos ⁡ 2 θ \frac { u }{ 2 } \sqrt { 2+\cos ^{ 2 }{ \theta } } u 2 1 + 3 cos ⁡ 2 θ \frac { u }{ 2 } \sqrt { 1+3\cos ^{ 2 }{ \theta } }

Swapnil Das by Swapnil Das , 20, India

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1 solution

Tan LeTian
Feb 27, 2016

-horizontal components of the velocity of the patricle is always constant, which is u cos ⁡ θ u\cos\theta

-vertical component of velocity decrease uniformly due to the gravitational acceleration

-Since the vertical component of velocity decrease linearly, we can find average vertical velocity by
(Final vertical velocity +initial vertical velocity )/2
= ( 0 + u sin ⁡ θ ) / 2 =(0+u\sin\theta)/2
= 1 2 u sin ⁡ θ =\frac{1}{2}u\sin\theta

To find average resultant velocity use Pythagoras theorem, Average resultant velocity = [ 1 2 u sin ⁡ θ ] ² + [ u cos ⁡ θ ] ² =\sqrt{ [\frac{1}{2}u\sin\theta]² +[u\cos\theta]²} = 1 2 u 1 + 3 cos ⁡ ² θ =\frac{1}{2}u \sqrt{1+3 \cos²\theta}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Please use correct Latex formatting for clarity, thanks!

Swapnil Das - 5 years, 3 months ago

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Just made some editing , first time posting a solution ,still learning , my apologise if there are any mistakes

Tan LeTian - 5 years, 3 months ago

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No need to apologize, we are friends indeed. Thank you :)

Swapnil Das - 5 years, 3 months ago

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