A particle is projected from the ground with an initial speed of `5 m s^(-1)` at an angle of projection 60° with horizontal. The average velocity of the particle between its time of projection and the time it reaches highest point of trajectory

Options

A. `1.25 ms^(-1)`
B. `1 ms^(-1)`
C. `1.25 sqrt7ms^(-1)`
D. `2 ms^(-1)`

Answer

To find the average velocity of a particle projected from the ground with an initial speed of \(5 \, \text{m/s}\) at an angle of \(60^\circ\) with the horizontal, we can follow these steps:

### Step 1: Resolve the initial velocity into components
The initial velocity \(u\) can be resolved into horizontal and vertical components using trigonometric functions.

- Horizontal component \(u_x = u \cos \theta\)
- Vertical component \(u_y = u \sin \theta\)

Given:
- \(u = 5 \, \text{m/s}\)
- \(\theta = 60^\circ\)

Calculating the components:
\[ u_x = 5 \cos 60^\circ = 5 \times \frac{1}{2} = 2.5 \, \text{m/s} \]
\[ u_y = 5 \sin 60^\circ = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \]

### Step 2: Determine the time to reach the highest point
At the highest point of the trajectory, the vertical component of the velocity becomes zero. We can use the following kinematic equation to find the time \(t\) to reach the highest point:
\[ v_y = u_y - g t \]
Where:
- \(v_y = 0\) (velocity at the highest point)
- \(g = 9.81 \, \text{m/s}^2\) (acceleration due to gravity)

Setting \(v_y = 0\):
\[ 0 = \frac{5\sqrt{3}}{2} - 9.81 t \]
Solving for \(t\):
\[ 9.81 t = \frac{5\sqrt{3}}{2} \]
\[ t = \frac{5\sqrt{3}}{2 \times 9.81} \approx 0.433 \, \text{s} \]

### Step 3: Calculate the displacement at the highest point
The horizontal displacement \(x\) at time \(t\) is given by:
\[ x = u_x \cdot t \]
Substituting the values:
\[ x = 2.5 \times 0.433 \approx 1.083 \, \text{m} \]

The vertical displacement \(y\) at time \(t\) can be calculated using:
\[ y = u_y t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ y = \frac{5\sqrt{3}}{2} \times 0.433 - \frac{1}{2} \times 9.81 \times (0.433)^2 \]
Calculating \(y\):
\[ y \approx \frac{5\sqrt{3}}{2} \times 0.433 - \frac{1}{2} \times 9.81 \times 0.187 \]
\[ y \approx 1.118 - 0.919 \approx 0.199 \, \text{m} \]

### Step 4: Calculate the average velocity
The average velocity \(\vec{V}_{avg}\) from the point of projection to the highest point is given by:
\[ \vec{V}_{avg} = \frac{\Delta \vec{r}}{\Delta t} \]
Where \(\Delta \vec{r} = (x, y)\) and \(\Delta t = t\).

Calculating the average velocity:
\[ \vec{V}_{avg} = \frac{(1.083 \hat{i} + 0.199 \hat{j})}{0.433} \]
\[ \vec{V}_{avg} = \left(\frac{1.083}{0.433} \hat{i} + \frac{0.199}{0.433} \hat{j}\right) \approx (2.5 \hat{i} + 0.46 \hat{j}) \, \text{m/s} \]

### Step 5: Calculate the magnitude of the average velocity
The magnitude of the average velocity is given by:
\[ |\vec{V}_{avg}| = \sqrt{(V_{avg_x})^2 + (V_{avg_y})^2} \]
Substituting the values:
\[ |\vec{V}_{avg}| = \sqrt{(2.5)^2 + (0.46)^2} \approx \sqrt{6.25 + 0.2116} \approx \sqrt{6.4616} \approx 2.54 \, \text{m/s} \]

### Final Answer
The average velocity of the particle between its time of projection and the time it reaches the highest point of trajectory is approximately \(2.54 \, \text{m/s}\).

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