Aball is thrown upward from the top of a 25.0 m tall building. the ball’s initial speed is 12.0 m/sec. at the same instant, a person is running on the ground at a distance of 31.0 m from the building. what must be the average speed of the person if he is to catch the ball at the bottom of the building?

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5at²

Initial velocity, u = 12 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -25 m

Substituting

             -25 = 12 x t + 0.5 x -9.81 x t²

               4.905 t² -12t - 25 = 0

              t = 3.79 sec

Ball hits ground after 3.79 seconds.

So person need to cover 31 m in 3.79 seconds

Consider the horizontal motion of person

We have equation of motion s = ut + 0.5at²

Initial velocity, u = ?

Acceleration, a = 0 m/s²

Displacement, s = 31 m

Time, t = 3.79 seconds

Substituting

             31 = u x 3.79 + 0.5 x 0 x 3.71²

               u = 8.18 m/s

Person must have 8.18 m/s to catch the ball

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