A ball is thrown from the ground with a velocity of `20sqrt3` m/s making an angle of `60^@` with the horizontal. The ball will be at a height of 40 m from the ground after a time t equal to `(g=10 ms^(-2))`

Options

A. `sqrt2` sec
B. `sqrt3` sec
C. 2 sec
D. 3 sec

Explanation

To solve the problem step by step, we will use the equations of motion for projectile motion.

### Step 1: Identify the given data
- Initial velocity \( u = 20\sqrt{3} \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)
- Height \( h = 40 \, \text{m} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the vertical component of the initial velocity
The vertical component of the initial velocity \( u_y \) can be calculated using the formula:
\[ u_y = u \sin \theta \]
Substituting the values:
\[ u_y = 20\sqrt{3} \cdot \sin(60^\circ) = 20\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 30 \, \text{m/s} \]

### Step 3: Use the equation of motion to find the time at which the ball reaches 40 m height
The vertical motion can be described by the equation:
\[ h = u_y t - \frac{1}{2} g t^2 \]
Substituting the known values:
\[ 40 = 30t - \frac{1}{2} \cdot 10 \cdot t^2 \]
This simplifies to:
\[ 40 = 30t - 5t^2 \]
Rearranging gives us a standard quadratic equation:
\[ 5t^2 - 30t + 40 = 0 \]

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5 \), \( b = -30 \), and \( c = 40 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-30)^2 - 4 \cdot 5 \cdot 40 = 900 - 800 = 100 \]
Now substituting into the quadratic formula:
\[ t = \frac{30 \pm \sqrt{100}}{2 \cdot 5} = \frac{30 \pm 10}{10} \]
This gives us two possible solutions for \( t \):
\[ t_1 = \frac{40}{10} = 4 \, \text{s} \]
\[ t_2 = \frac{20}{10} = 2 \, \text{s} \]

### Step 5: Interpret the results
The ball reaches a height of 40 m at two different times:
- At \( t = 2 \, \text{s} \) while going up.
- At \( t = 4 \, \text{s} \) while coming down.

### Final Answer
The ball will be at a height of 40 m from the ground at \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \).

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