(give a numerical answer, 50 points) a projectile launcher is placed on the floor and pointed straight up (90 degrees), the projectile (ball) leaves the launcher 27 cm above the floor. the ball has a mass of 6.58 g. we measure the maximum height above the floor reached by the ball. we do this three times and get the following results all measured in meters.: 2.32, 2.26 and 2.37. find the speed at which the ball leaves the launcher.

 Speed at which the ball leaves the launcher = 6.34 m/s

Explanation:

  The readings of height above the floor reached by the ball = 2.32, 2.26 and 2.37

  Mean value

  Initial height of launcher = 27 cm = 0.27 cm

  So maximum height of projection = 2.317-0.27 = 2.047 meter

  We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

 In this case we need to consider only vertical factors.

      Vertical displacement = 2.047 m, acceleration = acceleration due to gravity = , final velocity = 0 m/s, we need to calculate initial velocity.

    So, speed at which the ball leaves the launcher = 6.34 m/s

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray

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