A ball of mass 400 gm is dropped from tower of height 90 m. Find: i) Time to reach the ground. ii) Velocity with which it strike the ground and its final momentum.

$$
h = ut + \frac{1}{2}gt^2
$$

$$
90 \text{ m} = (0 \text{ m/s})t + \frac{1}{2}(9.8 \text{ m/s}^2)t^2
$$
$$
90 = 4.9t^2
$$

$$
t^2 = \frac{90}{4.9}
$$
$$
t^2 \approx 18.3673 \text{ s}^2
$$
$$
t = \sqrt{18.3673}
$$
$$
t \approx 4.286 \text{ s}
$$
The time to reach the ground is $\boxed{\text{4.29 s}}$.

$$
v^2 = u^2 + 2gh
$$

$$
v^2 = (0 \text{ m/s})^2 + 2(9.8 \text{ m/s}^2)(90 \text{ m})
$$
$$
v^2 = 0 + 1764 \text{ m}^2/\text{s}^2
$$
$$
v = \sqrt{1764}
$$
$$
v = 42 \text{ m/s}
$$
The velocity with which it strikes the ground is $\boxed{\text{42 m/s}}$.

$$
p = mv
$$

$$
p = (0.4 \text{ kg})(42 \text{ m/s})
$$
$$
p = 16.8 \text{ kg m/s}
$$
The final momentum is $\boxed{\text{16.8 kg m/s}}$.