Asked by Jake
                5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 22.0 m/s when it reaches a maximum height of 9.0 m above the ground. 
What is the speed of the ball when it leaves Sarah's hand?
            
        What is the speed of the ball when it leaves Sarah's hand?
Answers
                    Answered by
            Steve
            
    h(t) = -4.9t^2 + vt + 1.5
max height reached when t = v/9.8
9.0 = -4.9(v^2/4.9^2) + v(v/4.9) + 1.5
v = 7√3 = 12.1244 m/s
That is the vertical component. So, the speed is
√(7√3)^2 + 22^2 = 25.12 m/s
    
max height reached when t = v/9.8
9.0 = -4.9(v^2/4.9^2) + v(v/4.9) + 1.5
v = 7√3 = 12.1244 m/s
That is the vertical component. So, the speed is
√(7√3)^2 + 22^2 = 25.12 m/s
                    Answered by
            Evan
            
    While Steve is mostly right there is an error in his algebra. The third line should read: 
9.0=-4.9(v^2/9.8^2)+v(v/9.8)+1.5.
The rest can be correctly solved using this equation.
    
9.0=-4.9(v^2/9.8^2)+v(v/9.8)+1.5.
The rest can be correctly solved using this equation.
                    Answered by
            Jasmeet Singh
            
    2*a(x-x0)=(v^2-v0^2)
2*-9.8(9-1.5)=(v^2-(22*22))
v=18.35
    
2*-9.8(9-1.5)=(v^2-(22*22))
v=18.35
                    Answered by
            David 
            
    Where does the t=v/9.8 come form 
    
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