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 use the equations of projectile motion.

### Given:
- Time of flight (T) = 10 seconds
- Horizontal distance (Range, R) = 500 m
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 1: Find the Horizontal and Vertical Components of Velocity

The horizontal range (R) of a projectile is given by the formula:
\[ R = u \cos(\theta) \cdot T \]
Where:
- \( u \) = initial velocity
- \( \theta \) = angle of projection
- \( T \) = time of flight

Rearranging the formula to find \( u \cos(\theta) \):
\[ u \cos(\theta) = \frac{R}{T} = \frac{500 \, \text{m}}{10 \, \text{s}} = 50 \, \text{m/s} \]

### Step 2: Find the Vertical Component of Velocity

The time of flight for a projectile is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]

From this, we can express \( u \sin(\theta) \):
\[ u \sin(\theta) = \frac{gT}{2} = \frac{9.8 \, \text{m/s}^2 \cdot 10 \, \text{s}}{2} = 49 \, \text{m/s} \]

### Step 3: Find the Angle of Projection

Now, we have two equations:
1. \( u \cos(\theta) = 50 \, \text{m/s} \)
2. \( u \sin(\theta) = 49 \, \text{m/s} \)

To find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{u \sin(\theta)}{u \cos(\theta)} = \frac{49}{50} \]

Now, we can find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{49}{50}\right) \]

Calculating this gives:
\[ \theta \approx 44.42^\circ \]

### Step 4: Find the Velocity of Projection

Now we can find the initial velocity \( u \) using the Pythagorean theorem:
\[ u = \sqrt{(u \cos(\theta))^2 + (u \sin(\theta))^2} \]
\[ u = \sqrt{(50)^2 + (49)^2} = \sqrt{2500 + 2401} = \sqrt{4901} \]

Calculating this gives:
\[ u \approx 70 \, \text{m/s} \]

### Step 5: Find the Velocity of the Ball after 5 seconds

After 5 seconds, the vertical component of the velocity will be:
\[ v_y = u \sin(\theta) - g t = 49 \, \text{m/s} - 9.8 \, \text{m/s}^2 \cdot 5 \, \text{s} \]
\[ v_y = 49 \, \text{m/s} - 49 \, \text{m/s} = 0 \, \text{m/s} \]

The horizontal component of the velocity remains constant:
\[ v_x = u \cos(\theta) = 50 \, \text{m/s} \]

Thus, the resultant velocity after 5 seconds is:
\[ v = \sqrt{(v_x)^2 + (v_y)^2} = \sqrt{(50)^2 + (0)^2} = 50 \, \text{m/s} \]

### Summary of Results:
(i) Angle of projection \( \theta \approx 44.42^\circ \)
(ii) Velocity of projection \( u \approx 70 \, \text{m/s} \)
(iii) Velocity of the ball after 5 seconds \( v \approx 50 \, \text{m/s} \)