A hunter is riding an elephant of height `4 m` moving in straight line with uniform speed of `2m//sec`. A deer running with a speed `V` in front at a distance of `4sqrt(5)` moving perpendicular to the direction of motion of the elephant. If hunter can throw his spear with a speed of `10m//sec`. relative to the elephant, then at what angle `theta` to it's direction of motion must he throw his spear horizontally for a successful hit. Find also the speed 'V' of the deer.

Answer

To solve the problem step by step, we will analyze the motion of the hunter, the elephant, and the deer.

### Step 1: Understanding the scenario
The elephant is moving with a speed of \(2 \, \text{m/s}\) in a straight line, and the hunter is at a height of \(4 \, \text{m}\). The deer is running perpendicular to the direction of the elephant's motion at a distance of \(4\sqrt{5} \, \text{m}\). The spear is thrown at an angle \(\theta\) with respect to the direction of motion of the elephant.

### Step 2: Calculate the time of flight of the spear
The spear is thrown from a height of \(4 \, \text{m}\). The vertical motion of the spear can be described by the equation of motion:
\[ h = \frac{1}{2} g t^2 \]
where \(h = 4 \, \text{m}\) and \(g = 9.8 \, \text{m/s}^2\). Rearranging gives:
\[ 4 = \frac{1}{2} \cdot 9.8 \cdot t^2 \]
\[ t^2 = \frac{8}{9.8} \approx 0.8163 \]
\[ t \approx \sqrt{0.8163} \approx 0.9 \, \text{s} \]

### Step 3: Determine the horizontal distance covered by the spear
The horizontal component of the spear's velocity can be expressed as:
\[ v_x = v_{elephant} + v_{spear} \cdot \cos(\theta) \]
where \(v_{elephant} = 2 \, \text{m/s}\) and \(v_{spear} = 10 \, \text{m/s}\). The horizontal distance the spear must cover to hit the deer is \(4\sqrt{5} \, \text{m}\).

### Step 4: Set up the equation for horizontal distance
The horizontal distance covered by the spear in time \(t\) is given by:
\[ \text{Distance} = v_x \cdot t \]
Substituting the known values:
\[ 4\sqrt{5} = (2 + 10 \cos(\theta)) \cdot 0.9 \]
Solving for \(\cos(\theta)\):
\[ 4\sqrt{5} = (2 + 10 \cos(\theta)) \cdot 0.9 \]
\[ \frac{4\sqrt{5}}{0.9} = 2 + 10 \cos(\theta) \]
Calculating \(4\sqrt{5} \approx 8.944\):
\[ \frac{8.944}{0.9} \approx 9.93 \]
\[ 9.93 = 2 + 10 \cos(\theta) \]
\[ 10 \cos(\theta) = 9.93 - 2 = 7.93 \]
\[ \cos(\theta) = \frac{7.93}{10} = 0.793 \]

### Step 5: Calculate the angle \(\theta\)
Using the cosine value:
\[ \theta = \cos^{-1}(0.793) \approx 37^\circ \]

### Step 6: Find the speed \(V\) of the deer
The vertical component of the spear's velocity at the time of hitting the deer must equal the speed of the deer:
\[ V = v_{spear} \cdot \sin(\theta) \]
Calculating \(\sin(37^\circ) \approx 0.6\):
\[ V = 10 \cdot 0.6 = 6 \, \text{m/s} \]

### Final Answers:
- The angle \(\theta\) at which the spear must be thrown is approximately \(37^\circ\).
- The speed \(V\) of the deer is \(6 \, \text{m/s}\).