Find the Launch Speed for a 20m High Projectile

AI Thread Summary
To find the launch speed of a projectile that reaches a maximum height of 20 m, the vertical and horizontal components of velocity are analyzed using the equations for projectile motion. The direction of the velocity 1.0 second after launch is given as 20°, which helps in determining the initial vertical component of velocity. After several attempts, the solution was found to be a launch speed of 34 m/s at an angle of 36°. The discussion highlights the collaborative effort in problem-solving and the importance of perseverance in tackling physics problems. The thread serves as a resource for understanding projectile motion calculations.
voygehr
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Homework Statement


The maximum height reached by a projectile is 20 m. The direction of the velocity 1.0 s after launch is 20°. Find the speed of launch.


Homework Equations


vertical component of velocity = u \sin\theta - gt
horizontal component of velocity = u\cos\theta
where theta is the launch angle


The Attempt at a Solution


I made several attempts to solve this but I can't seem to figure it out. Are we presented with enough information?
 
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welcome to pf!

hi voygehr! welcome to pf!

(have a theta: θ :wink:)

find vy0 first …

show us what you get :smile:​
 
just solved it!

v=34 m/s at 36°

thanks anyway! (do I delete this thread now?)
 
no, it remains for ever as a triumph of human perseverance and ingenuity! :biggrin:
 
haha excellent!
 

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