If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is
v = -g t
where g = 9.80 m/s² is the magnitude of acceleration due to gravity.
Its height y is
y = 40 m - 1/2 g t²
The ball is dropped from a 40 m height, so that it takes
0 = 40 m - 1/2 g t²
==> t = √(80/g) s ≈ 2.86 s
for it to reach the ground, after which time it attains a velocity of
v = -g (√(80/g) s)
==> v = -√(80g) m/s ≈ -28.0 m/s
During the next bounce, the ball's speed is halved, so its height is given by
y = (14 m/s) t - 1/2 g t²
Solve y = 0 for t to see how long it's airborne during this bounce:
0 = (14 m/s) t - 1/2 g t²
0 = t (14 m/s - 1/2 g t)
==> t = 28/g s ≈ 2.86 s
So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of
y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²
==> (i) y ≈ 7.5 m
(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.
During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is
y = (3.5 m/s) t - 1/2 g t²
Solve y = 0 for t :
0 = (3.5 m/s) t - 1/2 g t²
0 = t (3.5 m/s - 1/2 g t)
==> (iii) t = 7/g m/s ≈ 0.714 s
As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be
v = 7 m/s - g t
At 6 s, the ball has velocity
(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s