A projectile is thrown with a velocity `vec v _(0) = 3hati + 4 hatj ms ^(-1)` where `hat I and hatj` are the unit vectors along the horizontal and vertical directions respectively. The speed of the projectile at the highest point of its motion in `ms ^(-1)` is

Options

A. 3
B. 4
C. 5
D. `0.75`

Answer

To find the speed of the projectile at the highest point of its motion, we can follow these steps:

### Step 1: Identify the initial velocity components
The initial velocity of the projectile is given as:
\[ \vec{v_0} = 3 \hat{i} + 4 \hat{j} \, \text{ms}^{-1} \]
From this, we can identify the horizontal and vertical components of the initial velocity:
- \( u_x = 3 \, \text{ms}^{-1} \) (horizontal component)
- \( u_y = 4 \, \text{ms}^{-1} \) (vertical component)

### Step 2: Determine the angle of projection
We can find the angle of projection \( \theta \) using the tangent function:
\[ \tan \theta = \frac{u_y}{u_x} = \frac{4}{3} \]
This gives us the angle of projection, but we do not need to calculate the angle explicitly for this problem.

### Step 3: Analyze the motion at the highest point
At the highest point of its trajectory, the vertical component of the velocity becomes zero because the projectile momentarily stops rising before falling back down. Therefore:
\[ v_y = 0 \, \text{ms}^{-1} \]

### Step 4: Calculate the horizontal component of the velocity
The horizontal component of the velocity remains constant throughout the projectile's motion since there is no horizontal acceleration (assuming no air resistance). Thus:
\[ v_x = u_x = 3 \, \text{ms}^{-1} \]

### Step 5: Calculate the speed at the highest point
The speed of the projectile at the highest point is given by the magnitude of the velocity vector, which in this case only has the horizontal component:
\[ \text{Speed} = \sqrt{v_x^2 + v_y^2} = \sqrt{(3)^2 + (0)^2} = \sqrt{9} = 3 \, \text{ms}^{-1} \]

### Final Answer
The speed of the projectile at the highest point of its motion is:
\[ \text{Speed} = 3 \, \text{ms}^{-1} \]

---