An arcade ball is thrown with an initial speed of 7.0 m/s and follows the trajectory shown. The ball enter the basket .95 seconds after it is launched. What are the distances x and y?

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Complete Question

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Answer:

[tex]x = 4.70 \ m[/tex]     ,   [tex]y = 0.2803 \ m[/tex]

Explanation:

From the question we are told that

     The initial velocity of the arcade ball is  [tex]u = 7.0 \ m/s[/tex]

     The time taken by the ball to enter the basket is  [tex]t = 0.95 \ s[/tex]

Generally the x -component of the initial  velocity is  

            [tex]u_x = 7 * cos (45)[/tex]    

=>         [tex]u_x = 4.95 \ m/s[/tex]

Generally the y -component of the initial  velocity is  

            [tex]u_x = 7 * sin(45)[/tex]    

=>         [tex]u_y = 4.95 \ m/s[/tex]

  Generally the distance x is mathematically represented as

                 [tex]x = u_x * t[/tex]

=>              [tex]x = 4.95 * 0.95[/tex]      

=>              [tex]x = 4.70 \ m[/tex]    

Generally from kinematic equation the distance y is mathematically represented as

               [tex]y = u_y * t - \frac{1}{2} * g * t^2[/tex]

=>            [tex]y = 4.95 * 0.95 - \frac{1}{2} * 9.8 * 0.95^2[/tex]

=>            [tex]y = 0.2803 \ m[/tex]

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