A projectile's launch speed is five times its speed at maximum height.Find launch angle.How do I find this?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

3
Mar 11, 2018

Answer:

#cos^-1(1/5) ~~ 78^circ#

Explanation:

If the angle of projection is #alpha# and the projection speed is #u#, the projectile starts off with the horizontal and vertical velocity components #u cos alpha# and #u sin alpha#, respectively. Of these, the horizontal component of the velocity is constant throughout the motion, while the vertical component keeps on changing at a constant rate #-g#.

At the topmost point of the path, the projectile's vertical velocity component vanishes - and thus its velocity at that point is exactly horizontal, and its magnitude is #u cos alpha#.

Thus $u = 5u cos alpha implies cos alpha = 1/5#

Was this helpful? Let the contributor know!
1500