What Is the Launch Angle of a Ball Thrown Against a Wall?

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SUMMARY

The problem involves calculating the launch angle of a ball thrown with an initial velocity of 20 m/s towards a wall 50 meters away, starting from a height of 1.8 meters and hitting the wall at a height of 1.2 meters. The key equations used are 50 = 20t*cos(theta) and -0.6 = 20t*sin(theta) - 0.5(9.8)(t^2). By rearranging the first equation to solve for time (t) and substituting it into the second equation, the angle (theta) can be determined without directly calculating the time.

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Homework Statement


A ball is thrown with initial velocity 20m/sec against a wall 50 meters away. It is released from a height of 1.8 meters, and hits the wall at 1.2 meters. Find the angle of the throw.


Homework Equations


50=20t*cos(theta)
-.6=20t*sin(theta)-.5(9.8)(t^2)


The Attempt at a Solution


Is it possible to solve this problem without knowing t(time between launch and hitting the wall)? Thanks for the help, I need to know if I'm missing a formula or something.
 
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im not 100% sure about this but you don't need to find t, you just need to rearange your 1st equatoin so you have t=... and then substitute that into your 2nd equation and then you can find theta. maybe
 

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