Question

(Please 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.

Answer 1

Answer:

 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 $=\frac{2.32+2.26+2.37}{3} =2.317 m$

  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, $v^2=u^2+2as$, 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 = $-9.81m/s^2$, final velocity = 0 m/s, we need to calculate initial velocity.

     $0^2=u^2-2*9.81*2.047\\ \\ u=\sqrt{2*9.81*2.047} \\ \\ u=6.34 m/s$

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