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 two parts: finding the speed and angle of projection as seen from the truck and then as seen from the road.

### Step 1: Calculate the time taken for the truck to travel 58.8 m

The truck moves with a constant velocity of \( v_t = 14.7 \, \text{m/s} \). The distance traveled by the truck is \( d = 58.8 \, \text{m} \).

Using the formula for time:
\[ t = \frac{d}{v_t} \]
Substituting the values:
\[ t = \frac{58.8 \, \text{m}}{14.7 \, \text{m/s}} = 4 \, \text{s} \]

### Step 2: Determine the speed of projection as seen from the truck

Since the ball returns to the person’s hand after 4 seconds, it takes 2 seconds to reach the maximum height and another 2 seconds to come back down.

Using the equation of motion:
\[ v = u + at \]
At the maximum height, the final velocity \( v = 0 \). The acceleration due to gravity \( a = -g \) (taking downward as negative).

Thus, we have:
\[ 0 = u - g \cdot t \]
For the upward journey, \( t = 2 \, \text{s} \):
\[ 0 = u - g \cdot 2 \]
Rearranging gives:
\[ u = g \cdot 2 \]
Substituting \( g \approx 9.8 \, \text{m/s}^2 \):
\[ u = 9.8 \cdot 2 = 19.6 \, \text{m/s} \]

### Step 3: Determine the angle of projection as seen from the truck

Since the ball is thrown vertically upward, the angle of projection \( \theta \) is:
\[ \theta = 90^\circ \]

### Step 4: Calculate the speed of projection as seen from the road

To find the speed of the ball as observed from the road, we need to consider the horizontal component of the velocity. The horizontal velocity of the truck is \( v_t = 14.7 \, \text{m/s} \) and the vertical velocity of the ball is \( u = 19.6 \, \text{m/s} \).

Using Pythagoras theorem to find the resultant velocity \( v \):
\[ v = \sqrt{(v_t)^2 + (u)^2} \]
Substituting the values:
\[ v = \sqrt{(14.7)^2 + (19.6)^2} \]
Calculating:
\[ v = \sqrt{216.09 + 384.16} = \sqrt{600.25} \approx 24.5 \, \text{m/s} \]

### Step 5: Determine the angle of projection as seen from the road

The angle \( \theta' \) can be found using the tangent function:
\[ \tan(\theta') = \frac{u}{v_t} \]
Substituting the values:
\[ \tan(\theta') = \frac{19.6}{14.7} \]
Calculating:
\[ \theta' = \tan^{-1}\left(\frac{19.6}{14.7}\right) \approx 53.1^\circ \]

### Final Results

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

- **As seen from the road:**
- Speed of projection \( v \approx 24.5 \, \text{m/s} \)
- Angle of projection \( \theta' \approx 53.1^\circ \)