Question

A ball is thrown horizontally from the top of a building 30.2 m high. The ball strikes the ground at a point 94.7 m from the base of the building. The acceleration of gravity is 9.8 m/s 2 . Find the time the ball is in motion. Answer in units of s. 015 (part 2 of 4) 10.0 points Find the initial velocity of the ball. Answer in units of m/s. 016 (part 3 of 4) 10.0 points Find the x component of its velocity just before it strikes the ground. Answer in units of m/s. 017 (part 4 of 4) 10.0 points Find the y component of its velocity just before it strikes the ground. Answer in units of m/s.

Answer 1

Answer:

1) The ball is in motion for 2.48 seconds.

2) Initial velocity of ball is 38.16 m/s

3)  x component of velocity before hitting the ground is 38.16 m/s  

4) y component of velocity before hitting the ground is 24.33 m/s  

Explanation:

1) Consider the vertical motion of ball

          Displacement, s =  30.2 m

          Initial velocity, u = 0 m/s

          Acceleration, a = 9.81 m/s²

 Substituting in s = ut + 0.5 at²

         30.2 = 0 x t + 0.5  x 9.81 x t²

           t = 2.48 s

    The ball is in motion for 6.16 seconds.        

2) Consider the horizontal motion of ball

          Displacement, s =  94.7 m

          Time, t = 2.48 s

          Acceleration, a = 0 m/s²

 Substituting in s = ut + 0.5 at²

         94.7 = u x 2.48 + 0.5  x 0 x 2.48²

           u = 38.16 m/s

    Initial velocity of ball = 38.16 m/s    

3) x component of velocity will not change since acceleration is zero along x direction.

       x component of velocity before hitting the ground = 38.16 m/s  

4) Consider the vertical motion of ball

          Initial velocity, u = 0 m/s

          Acceleration, a = 9.81 m/s²

          Time, t = 2.48 s

 Substituting in v = u + at

         v = 0 + 9.81 x 2.48

           v = 24.33 m/s

    y component of velocity before hitting the ground = 24.33 m/s