Question

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?

Answer 1

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

$x = 4.70 \ m$     ,   $y = 0.2803 \ m$

Explanation:

From the question we are told that

     The initial velocity of the arcade ball is  $u = 7.0 \ m/s$

     The time taken by the ball to enter the basket is  $t = 0.95 \ s$

Generally the x -component of the initial  velocity is  

            $u_x = 7 * cos (45)$    

=>         $u_x = 4.95 \ m/s$

Generally the y -component of the initial  velocity is  

            $u_x = 7 * sin(45)$    

=>         $u_y = 4.95 \ m/s$

  Generally the distance x is mathematically represented as

                 $x = u_x * t$

=>              $x = 4.95 * 0.95$      

=>              $x = 4.70 \ m$    

Generally from kinematic equation the distance y is mathematically represented as

               $y = u_y * t - \frac{1}{2} * g * t^2$

=>            $y = 4.95 * 0.95 - \frac{1}{2} * 9.8 * 0.95^2$

=>            $y = 0.2803 \ m$