Solving for Angle of Vector: Ball Bouncing Problem | Physics Homework Help

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SUMMARY

The discussion centers on calculating the angle of a vector in a ball bouncing problem, where the ball reaches a height of 3.10 m and lands 0.70 m away after the first bounce. The vertical velocity (Vy) is calculated as 7.799 m/s using the equation H = 0.5(v^2/g). The horizontal velocity (Vx) is determined to be 0.441 m/s. The correct angle is derived from the tangent function, resulting in an angle of approximately 79.4 degrees, which was initially miscalculated.

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Homework Statement



A ball is tossed so that it bounces off the ground, rises to a height of 3.10 m, and then hits the ground again 0.70 m away from the first bounce.

Homework Equations


H= .5(v^2/g) for y velocity.


The Attempt at a Solution


Vy=squareroot of(60.822) Vy= 7.799 m/s
Vx= .70 m/1.59 s= .441 m/s
inversetangent(Vy+Vx)= angle ------ is this correct?

i did the inversetangent and it came out to be about 79.4 degrees. but that wasn't the correct answer :(
 
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your Vx , Vy are correct but the angle is wrong.

[tex]\tan(\theta)=\frac{V_y}{V_x}=\frac{7.79}{0.44}[/tex]
 
thank you! :)
 

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