A person is standing on a truck moving with a constant velocity of 14.7 m/s o a hrozontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection. a. as seen from the truck b. as seen fromt the road.

Explanation

To solve the problem step by step, we will break it down into parts and calculate the speed and angle of projection as seen from both the truck and the road.

### Step 1: Determine the Time of Flight
The truck moves with a constant velocity of \( V_t = 14.7 \, \text{m/s} \) and travels a distance of \( R = 58.8 \, \text{m} \). We can find the time \( t \) it takes for the truck to travel this distance using the formula:

\[ t = \frac{R}{V_t} \]

Substituting the values:

\[ t = \frac{58.8 \, \text{m}}{14.7 \, \text{m/s}} = 4 \, \text{s} \]

### Step 2: Analyze Motion from the Truck's Frame of Reference
From the truck's perspective, the ball is thrown vertically upward and then comes back down. Since the total time of flight is 4 seconds, the time taken to reach the maximum height is half of that:

\[ t_{\text{up}} = \frac{t}{2} = \frac{4 \, \text{s}}{2} = 2 \, \text{s} \]

### Step 3: Calculate the Initial Vertical Velocity
Using the equation of motion, where the final velocity at the maximum height is 0, we can find the initial vertical velocity \( u_y \):

\[ v = u + at \]

Here, \( v = 0 \), \( a = -g = -9.8 \, \text{m/s}^2 \), and \( t = 2 \, \text{s} \):

\[ 0 = u_y - 9.8 \times 2 \]
\[ u_y = 9.8 \times 2 = 19.6 \, \text{m/s} \]

### Step 4: Determine the Speed and Angle of Projection as Seen from the Truck
From the truck's perspective, the ball is thrown straight up with a speed of \( 19.6 \, \text{m/s} \) and an angle of projection of \( 90^\circ \) (since it goes straight up).

**Answer:**
- Speed as seen from the truck: \( 19.6 \, \text{m/s} \)
- Angle of projection as seen from the truck: \( 90^\circ \)

### Step 5: Analyze Motion from the Road's Frame of Reference
Now, we need to find the angle of projection and the initial speed as seen from the road. The horizontal component of the velocity \( u_x \) is equal to the truck's speed:

\[ u_x = V_t = 14.7 \, \text{m/s} \]

The vertical component of the velocity \( u_y \) is \( 19.6 \, \text{m/s} \) (as calculated earlier).

### Step 6: Calculate the Angle of Projection
Using the tangent function to find the angle \( \theta \):

\[ \tan \theta = \frac{u_y}{u_x} = \frac{19.6}{14.7} \]

Calculating \( \tan \theta \):

\[ \tan \theta \approx 1.333 \]

Now, taking the inverse tangent to find \( \theta \):

\[ \theta = \tan^{-1}(1.333) \approx 53^\circ \]

### Step 7: Calculate the Magnitude of the Initial Velocity
To find the magnitude of the initial velocity \( U \):

\[ U = \sqrt{u_x^2 + u_y^2} \]

Calculating \( U \):

\[ U = \sqrt{(14.7)^2 + (19.6)^2} = \sqrt{216.09 + 384.16} = \sqrt{600.25} \approx 24.5 \, \text{m/s} \]

**Answer:**
- Speed as seen from the road: \( 24.5 \, \text{m/s} \)
- Angle of projection as seen from the road: \( 53^\circ \)

### Summary of Answers:
- Speed as seen from the truck: \( 19.6 \, \text{m/s} \)
- Angle of projection as seen from the truck: \( 90^\circ \)
- Speed as seen from the road: \( 24.5 \, \text{m/s} \)
- Angle of projection as seen from the road: \( 53^\circ \)