A ball is projected from ground in such a way that after 10 seconds of projection it lands on ground 500 m away from the point of projection. Find out :-
(i) angle of projection
(ii) velocity of projection
(iii) Velocity of ball after 5 seconds

Answer

To solve the problem step by step, we will follow the principles of projectile motion.

### Given:
- Total time of flight (T) = 10 seconds
- Horizontal range (R) = 500 m
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 1: Find the horizontal component of the velocity

The horizontal range (R) can be expressed as:
\[ R = u \cdot \cos(\theta) \cdot T \]

Where:
- \( u \) = initial velocity
- \( \theta \) = angle of projection

Substituting the known values:
\[ 500 = u \cdot \cos(\theta) \cdot 10 \]
\[ u \cdot \cos(\theta) = \frac{500}{10} = 50 \, \text{m/s} \quad \text{(1)} \]

### Step 2: Find the vertical component of the velocity

The time of flight (T) can also be expressed in terms of the vertical component of the initial velocity:
\[ T = \frac{2u \cdot \sin(\theta)}{g} \]

Rearranging gives:
\[ u \cdot \sin(\theta) = \frac{g \cdot T}{2} \]
Substituting the known values:
\[ u \cdot \sin(\theta) = \frac{9.8 \cdot 10}{2} = 49 \, \text{m/s} \quad \text{(2)} \]

### Step 3: Find the angle of projection

Now we have two equations:
1. \( u \cdot \cos(\theta) = 50 \)
2. \( u \cdot \sin(\theta) = 49 \)

Dividing equation (2) by equation (1):
\[ \frac{u \cdot \sin(\theta)}{u \cdot \cos(\theta)} = \frac{49}{50} \]
This simplifies to:
\[ \tan(\theta) = \frac{49}{50} \]

Now, calculate \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{49}{50}\right) \approx 44.42^\circ \]

### Step 4: Find the initial velocity

Now, we can find \( u \) using either equation (1) or (2). We will use equation (1):
\[ u \cdot \cos(44.42^\circ) = 50 \]

Calculating \( \cos(44.42^\circ) \):
\[ \cos(44.42^\circ) \approx 0.707 \]
So,
\[ u \cdot 0.707 = 50 \]
\[ u = \frac{50}{0.707} \approx 70.71 \, \text{m/s} \]

### Step 5: Find the velocity of the ball after 5 seconds

After 5 seconds, the ball is at the peak of its trajectory. The vertical component of the velocity at this point is 0 m/s (as it has reached the maximum height). The horizontal component remains constant.

Using equation (1):
\[ u \cdot \cos(44.42^\circ) = 50 \, \text{m/s} \]

Thus, the total velocity after 5 seconds is:
\[ \text{Velocity} = \sqrt{(u \cdot \cos(\theta))^2 + (0)^2} = 50 \, \text{m/s} \]

### Summary of Results:
(i) Angle of projection \( \theta \approx 44.42^\circ \)

(ii) Velocity of projection \( u \approx 70.71 \, \text{m/s} \)

(iii) Velocity of the ball after 5 seconds \( = 50 \, \text{m/s} \)

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