To solve the problem, we need to analyze the projectile motion of the object at a height of 15 m. The velocity at this height is given as \( \vec{v} = 10 \hat{i} + 10 \hat{j} \) m/s, where \( \hat{i} \) is the horizontal direction and \( \hat{j} \) is the vertical direction. The acceleration due to gravity \( g \) is given as \( 10 \, \text{m/s}^2 \).
### Step 1: Identify the components of the velocity
The velocity of the projectile at the height of 15 m can be broken down into its horizontal and vertical components:
- Horizontal component \( v_x = 10 \, \text{m/s} \)
- Vertical component \( v_y = 10 \, \text{m/s} \)
### Step 2: Relate the horizontal component to the initial velocity
Since there is no acceleration in the horizontal direction, the horizontal component of the initial velocity \( u \cos \theta \) remains constant throughout the motion:
\[
u \cos \theta = v_x = 10 \, \text{m/s}
\]
### Step 3: Use the vertical motion equation
For the vertical motion, we can use the third equation of motion:
\[
v_y^2 = u_y^2 - 2g h
\]
Where:
- \( v_y = 10 \, \text{m/s} \) (final vertical velocity)
- \( u_y = u \sin \theta \) (initial vertical velocity)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 15 \, \text{m} \) (height)
Substituting the known values:
\[
10^2 = (u \sin \theta)^2 - 2 \cdot 10 \cdot 15
\]
\[
100 = (u \sin \theta)^2 - 300
\]
\[
(u \sin \theta)^2 = 400
\]
\[
u \sin \theta = 20 \, \text{m/s}
\]
### Step 4: Calculate the angle of projection
Now we have both components of the initial velocity:
- \( u \cos \theta = 10 \)
- \( u \sin \theta = 20 \)
We can find the angle of projection \( \theta \) using the tangent function:
\[
\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{20}{10} = 2
\]
Thus,
\[
\theta = \tan^{-1}(2)
\]
### Step 5: Calculate the time of flight
The time of flight \( T \) can be calculated using the formula:
\[
T = \frac{2 u \sin \theta}{g}
\]
Substituting the known values:
\[
T = \frac{2 \cdot 20}{10} = 4 \, \text{s}
\]
### Step 6: Calculate the range
The range \( R \) can be calculated as:
\[
R = u_x \cdot T
\]
Where \( u_x = u \cos \theta = 10 \, \text{m/s} \):
\[
R = 10 \cdot 4 = 40 \, \text{m}
\]
### Step 7: Calculate the maximum height
The maximum height \( H \) can be calculated using:
\[
H = \frac{(u \sin \theta)^2}{2g}
\]
Substituting the known values:
\[
H = \frac{20^2}{2 \cdot 10} = \frac{400}{20} = 20 \, \text{m}
\]
### Summary of Results
- Angle of projection \( \theta = \tan^{-1}(2) \)
- Time of flight \( T = 4 \, \text{s} \)
- Range \( R = 40 \, \text{m} \)
- Maximum height \( H = 20 \, \text{m} \)
