A particle is projected with the speed of `10 sqrt(5) m//"sec"`. at an angle of `60^(@)` from the horizontal. The velocity of the projectile when it reaches the height of 10 m is (`g = 9.8 m//"sec"^(2)`)

Options

A. `4sqrt(19)m//"sec"`
B. `sqrt(179)m//"sec"`
C. `15m//"sec"`
D. `5sqrt(15)m//"sec"`

Answer

To solve the problem, we need to find the velocity of the projectile when it reaches a height of 10 m. We will break this down step by step.

### Step 1: Determine the initial velocity components
The initial velocity \( u \) is given as \( 10\sqrt{5} \, \text{m/s} \) and the angle of projection \( \theta \) is \( 60^\circ \).

- The initial velocity in the vertical direction \( u_y \) is given by:
\[ u_y = u \sin \theta = 10\sqrt{5} \sin 60^\circ = 10\sqrt{5} \cdot \frac{\sqrt{3}}{2} = 5\sqrt{15} \, \text{m/s} \]

- The initial velocity in the horizontal direction \( u_x \) is given by:
\[ u_x = u \cos \theta = 10\sqrt{5} \cos 60^\circ = 10\sqrt{5} \cdot \frac{1}{2} = 5\sqrt{5} \, \text{m/s} \]

### Step 2: Use the vertical motion equation to find \( v_y \)
We will use the kinematic equation for vertical motion:
\[ v_y^2 = u_y^2 + 2a_y s_y \]
where:
- \( v_y \) is the final vertical velocity,
- \( u_y \) is the initial vertical velocity,
- \( a_y = -g = -9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( s_y = 10 \, \text{m} \) (the height reached).

Substituting the values:
\[ v_y^2 = (5\sqrt{15})^2 + 2(-9.8)(10) \]
Calculating \( (5\sqrt{15})^2 \):
\[ (5\sqrt{15})^2 = 25 \cdot 15 = 375 \]
Now substituting this back into the equation:
\[ v_y^2 = 375 - 196 = 179 \]
Thus,
\[ v_y = \sqrt{179} \, \text{m/s} \]

### Step 3: Determine the horizontal velocity \( v_x \)
Since there is no horizontal acceleration, the horizontal component of velocity remains constant:
\[ v_x = u_x = 5\sqrt{5} \, \text{m/s} \]

### Step 4: Calculate the resultant velocity \( v \)
The resultant velocity \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} \]
Calculating \( v_x^2 \) and \( v_y^2 \):
\[ v_x^2 = (5\sqrt{5})^2 = 25 \cdot 5 = 125 \]
\[ v_y^2 = 179 \]
Now substituting these values:
\[ v = \sqrt{125 + 179} = \sqrt{304} \]
Thus,
\[ v \approx 17.43 \, \text{m/s} \]

### Final Answer
The velocity of the projectile when it reaches the height of 10 m is \( \sqrt{304} \, \text{m/s} \) or approximately \( 17.43 \, \text{m/s} \).

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