To solve the problem step by step, we will break it down into two parts as specified in the question.
### Given Data:
- Height of release, \( H = 400 \, \text{m} \)
- Horizontal velocity component, \( v_x = ay \) where \( a = \sqrt{5} \, \text{s}^{-1} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
### Part (a): Finding the Horizontal Drift of the Particle
1. **Determine the time taken to reach the ground:**
The vertical motion of the particle can be described by the equation:
\[
H = \frac{1}{2} g t^2
\]
Substituting the known values:
\[
400 = \frac{1}{2} \times 10 \times t^2
\]
Simplifying gives:
\[
400 = 5 t^2 \implies t^2 = \frac{400}{5} = 80 \implies t = \sqrt{80} = 8.94 \, \text{s}
\]
2. **Calculate the horizontal drift:**
The horizontal velocity \( v_x \) is given by:
\[
v_x = a y = \sqrt{5} y
\]
The horizontal drift \( x \) can be found by integrating the horizontal velocity over time:
\[
dx = v_x dt = \sqrt{5} y dt
\]
Since \( y = H - \frac{1}{2} g t^2 \), we can express \( y \) in terms of \( t \):
\[
y = 400 - 5t^2
\]
Now, substituting \( y \) into the equation for \( v_x \):
\[
v_x = \sqrt{5} (400 - 5t^2)
\]
To find the drift \( x \), we integrate:
\[
x = \int_0^t v_x dt = \int_0^{8.94} \sqrt{5} (400 - 5t^2) dt
\]
Solving this integral gives:
\[
x = \sqrt{5} \left[ 400t - \frac{5}{3}t^3 \right]_0^{8.94}
\]
Evaluating this at \( t = 8.94 \):
\[
x = \sqrt{5} \left[ 400 \times 8.94 - \frac{5}{3} \times (8.94)^3 \right]
\]
After calculating, we find:
\[
x \approx 2.66 \, \text{km}
\]
### Part (b): Finding the Speed with which the Particle Strikes the Ground
1. **Calculate the vertical component of the velocity when it strikes the ground:**
The vertical velocity \( v_y \) can be calculated using:
\[
v_y = g t = 10 \times 8.94 = 89.4 \, \text{m/s}
\]
2. **Calculate the horizontal component of the velocity at the moment of impact:**
Using the previously calculated \( v_x \):
\[
v_x = \sqrt{5} \times 400 = 400\sqrt{5} \, \text{m/s}
\]
3. **Calculate the resultant speed at impact:**
The resultant speed \( v \) is given by:
\[
v = \sqrt{v_x^2 + v_y^2}
\]
Substituting the values:
\[
v = \sqrt{(400\sqrt{5})^2 + (89.4)^2}
\]
After calculating, we find:
\[
v \approx 400.89 \, \text{m/s}
\]
### Final Answers:
(a) The horizontal drift of the particle when it strikes the ground is approximately \( 2.66 \, \text{km} \).
(b) The speed with which the particle strikes the ground is approximately \( 400.89 \, \text{m/s} \).
