A tennis ball is thrown vertically upward with an initial velocity of +7.7 m/s. What will the ball's velocity be when it returns to starting...

s = ut +(1/2) gt^2      (1),is the equation of motion of the tennis ball.

s is the vertical displacement at any time from the starting point, and u is the initial velocity of the tennis ball(=7.7m/s) and  g is the acceleration due to gravity(=9.8m/s^2) and t is the time of motion

The ball when reaches back at the starting point its net displacement is 0. So, using this in (1) we get the time to reach back the starting point.

0=7.7t-(1/2)(9.8)t^2

0=15.4t+0.8t^2

(t)(15.4-9.8t) = 0

t=0 or t=15.4/9.8 =1.5714 secs is the time for the ball to reach back to the starting point. t=0 is the starting time when s=0.

The ball takes 7.7/g secs to loose its velocity to zero , which is 0.7857 secs  which is half the total time duration from start to reach back to the same postion.

So, from 0 velocity at the highest point with acceleration due to gravity, g m/s^2 falling free back in 0.7857 secs it regains the velocity = u+gt =0+gt= 9.8*0.7857 = 7.7 m/s downward while reaching the ground.(See its starting velocity was equal in magnitude but upward in direction.)