A tennis ball "A" is released from rest down a 10.0 m long inclined ramp with a uniform acceleration of 5.0 m/s2 . Another tennis ball "B" is initially located at the same height as ball "A" right above the lower edge of the ramp. Ball "B" is thrown upward with some initial speed at the same instant as the release of ball "A". A) What was the initial velocity of ball "B" so that "A" and "B" reach the bottom of the ramp at the same time?

4.8 m/s

Explanation:

For tennis ball "A", we are told that;

Initial velocity; u = 0 m/s

Distance; h = 10 m

Acceleration; a = 5 m/s²

Thus, using Newton's equation of motion, we can find time(t).

h = ut + ½at²

10 = 0 + ½(5)t²

10 = ½(5t²)

t² = 2 × 10/5

t² = 4

t = √4

t = 2 seconds

Now, this time would be the same for ball B since we are told that they reach the bottom at the same time.

Now, since ball B is thrown upwards, it means it is thrown against gravity. Thus;

h = -ut + ½gt²

10 = -2u + ½(9.8)(2²)

Where u is initial velocity of Ball B.

Thus;

10 = -2u + 19.6

2u = 19.6 - 10

2u = 9.6

u = 9.6/2

u = 4.8 m/s

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