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

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To determine the launch angle of a ball thrown against a wall, the initial velocity is 20 m/s, and the ball travels 50 meters horizontally while being released from a height of 1.8 meters and hitting the wall at 1.2 meters. The equations of motion are set up, with the horizontal distance equation being 50 = 20t*cos(theta) and the vertical motion equation being -0.6 = 20t*sin(theta) - 0.5(9.8)(t^2). The discussion suggests that time (t) can be eliminated by rearranging the first equation to express t and substituting it into the second equation. This approach allows for the calculation of the launch angle (theta) without needing to know the time directly. The solution hinges on correctly manipulating the equations to isolate theta.
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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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