A child throws a tennis ball, from ground level, perfectly vertical in the air, with an initial velocity of

Question

100m/s.What is its velocity at 5.1 seconds? What is its velocity at 15.3 seconds?

Answer 1

    From the problem you can say that the objects moves in one-dimension only. Supposing there is no air drag or resisting force to make the problem more simple, the solution will be...

    At y coordinates ( that is the coordinate in the vertical axis) the force that acts on the object(tennis ball) will be the force of gravity towards downward (or towards the center of the earth).

        so,

                     f = ma

                    ma = - mg            

(here negative sign means the downward direction of the force of gravity)

                    m ( Vf - Vi ) / t    =  - mg

            

        thereforce,

                    Vf = - gt + Vi

a. if Vi = 100 m/s after 5.1 seconds

                    Vf = - (9.8)(5.1) m/s + 100 m/s

                    Vf = 50.02 m/s

b. if Vi = 100 m.s after 15.3 seconds

                    Vf = - (9.8)(15.3) + 100 m/s

                    Vf = - 49.94

  (here the negative sign implies that the object is now falling down to earth)

c. if he throws it in outer space with an initial velocity of 120 m/s horizontally at 20 seconds later the velocity is the same as 120 m/s. This is because at the outer space there is no force that may act to the object (technically speaking if we don't consider the quantum theory and since the object mass is incomparable to the mass of planets and stars this is a good assumption).

    Using the Law of inertia of Isaac Newton, which he said that "an object in motion will remain in motion unless acted upon by external force", we can conclude that the object will have a constant velocity at any time.  

Answer 2

Take vertically upward direction as positive.

Initial velocity u = 100 m/s

Acceleration a = -g = -9.8 m/s2

v = u + at = 100 - 9.8 t

At t = 5.1 s,

v = 100 - 9.8 * 5.1 = 50 m/s

i.e. 50 m/s vertically upward.

At v = 15.3 s,

v = 100 - 9.8 * 15.3 = -50 m/s

i.e. 50 m/s vertically downward

In outer space, there is no significance of terms like horizontally and vertically. But, in absence of any external force, the velocity will remain as 120 m/s.

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Answer 3

9.8m/s is the gravity constant acceleration so

100 - (5.1 * 9.8)= 50.02

100 - (15.3 * 9.8)=-49.94

but there cant be negative velocity so its 48.94 going down

Answer 4

gravity = -9.8m/s/s

5.1 * -9.8 = -49.98

100 + -49.98 = 50.02m/s

100 + (15.3 * -9.8) = -49.94m/s

Answer 5

Vf = Vi +at

a) vf  = vi+at

        = 100 + (-10) (5.1)

    vf  =  49

b)  vf = vi + at

        =100 + (-10)(15.3)