A projectile is projected from ground with initial velocity `vecu=u_(0)hati+v_(0)hatj`. If acceleration due to gragvity (g) is along the negative y-direction then find maximum displacement in x-direction.

Options

A. `(2‘mu_1 mu_2‘)/g`
B. `(‘mu_1^2)/g`
C. `(4‘mu_1 mu_2‘)/g`
D. `(‘mu_2^2)/g`

Answer

To solve the problem of finding the maximum displacement in the x-direction for a projectile launched with an initial velocity \( \vec{u} = u_0 \hat{i} + v_0 \hat{j} \) under the influence of gravity, we can follow these steps:

### Step 1: Understand the Components of Motion
The initial velocity has two components:
- Horizontal (x-direction): \( u_0 \)
- Vertical (y-direction): \( v_0 \)

The acceleration due to gravity acts in the negative y-direction, which means \( a_y = -g \) and \( a_x = 0 \).

### Step 2: Calculate the Time to Reach Maximum Height
At the maximum height, the vertical component of the velocity becomes zero. We can use the kinematic equation:
\[ v_y = u_y + a_y t \]
At maximum height, \( v_y = 0 \), so:
\[ 0 = v_0 - g \left(\frac{t}{2}\right) \]
Rearranging gives:
\[ g \left(\frac{t}{2}\right) = v_0 \quad \Rightarrow \quad t = \frac{2v_0}{g} \]
This \( t \) is the total time of flight to reach the maximum height, but we need the total time of flight, which is \( 2t \):
\[ T = \frac{2v_0}{g} \]

### Step 3: Calculate the Horizontal Displacement
Since there is no horizontal acceleration, the horizontal displacement (range) can be calculated using:
\[ \text{Range} = u_x \cdot T \]
Substituting the values:
\[ \text{Range} = u_0 \cdot T = u_0 \cdot \left(\frac{2v_0}{g}\right) \]
Thus, the maximum displacement in the x-direction is:
\[ \text{Range} = \frac{2u_0 v_0}{g} \]

### Final Answer
The maximum displacement in the x-direction (range) is:
\[ \text{Range} = \frac{2u_0 v_0}{g} \]

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