What will the ball's velocity be when it returns to its starting point if a ball is thrown vertically upward with an initial velocity of +7.5 m/s?

Here the ball falls down in the same vertical path. So it
acquires the  velocity  of equal manitude but in the opposite direction  while reaching
the starting point compared to its projected velocity vertically
upward.

The motion of the ball from the start is given
by:

V = u+at. where u is the starting velocity and v is the
final velocity, g is the scceleration due to gravity, s is the distance
travelled.

The thown up ball goes up for a time t,  till
its upward velocity v becomes zero due to  the downward  pull of acceleration due to
gravity.

So v = 0 = u+at = 7.5m/s-9.8t. Or t = 7.5/9.8
sec.

During this time the ball has reached height of  Ss=
ut -(1/2) at^2 = 7.5(7.5/9.8) - (1/2)(9.8)(7.5/9.8)^2 = (7.5)^2/19.6= 2.8699m
nearly.

Nowe the ball traces back with the same path and
same distance.

So we aply v^2-u^2= 2gs, where v = initial
velocity at the height  = 0 and v  = final velocity whils reaching starting point.. ans
s = (7.5)^2/(2g).

So v^2-u^2 =
2gs.

v^2 = 2gs , a u = 0.

v =
2g (7.5)^2/(2g)

v^2 = 7.5^2

v
= 7.5 m/s

So the velocity of the ball  while reaching back
the starting point is 7.5m/s downward.