A body is projected from the ground with some speed at some angle with the horizontal. Taking the horizontal and vertical direction to be x and y axis respectively and the point of projection as origin, calculate the minimum speed (in `ms^(-1)`) of projection so that it can pass through a point whose x and y coordinates are 30 m and 40 m respectively? Take `g=10ms^(-2)`

Answer

To solve the problem of finding the minimum speed of projection for a projectile to pass through the point (30 m, 40 m), we will follow these steps:

### Step 1: Understand the motion of the projectile
The horizontal and vertical motions of the projectile can be described by the following equations:
- Horizontal motion: \( x = u \cos \theta \cdot t \)
- Vertical motion: \( y = u \sin \theta \cdot t - \frac{1}{2} g t^2 \)

Where:
- \( u \) is the initial speed of projection
- \( \theta \) is the angle of projection
- \( g \) is the acceleration due to gravity (given as \( 10 \, \text{m/s}^2 \))
- \( t \) is the time of flight

### Step 2: Express time \( t \) in terms of \( x \)
From the horizontal motion equation, we can express time \( t \) as:
\[ t = \frac{x}{u \cos \theta} \]
Substituting \( x = 30 \, \text{m} \):
\[ t = \frac{30}{u \cos \theta} \]

### Step 3: Substitute \( t \) into the vertical motion equation
Now we substitute \( t \) into the vertical motion equation:
\[ y = u \sin \theta \cdot \left(\frac{30}{u \cos \theta}\right) - \frac{1}{2} g \left(\frac{30}{u \cos \theta}\right)^2 \]
Substituting \( y = 40 \, \text{m} \) and simplifying:
\[ 40 = 30 \tan \theta - \frac{1}{2} g \left(\frac{30^2}{u^2 \cos^2 \theta}\right) \]
Substituting \( g = 10 \):
\[ 40 = 30 \tan \theta - \frac{4500}{u^2 \cos^2 \theta} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ 30 \tan \theta - 40 = \frac{4500}{u^2 \cos^2 \theta} \]
Let \( \tan \theta = t \):
\[ 30t - 40 = \frac{4500}{u^2 (1 + t^2)} \]

### Step 5: Solve for \( u^2 \)
Rearranging gives:
\[ u^2 = \frac{4500}{(30t - 40)(1 + t^2)} \]

### Step 6: Find the minimum value of \( u^2 \)
To find the minimum speed, we need to differentiate \( u^2 \) with respect to \( t \) and set the derivative to zero. This will give us the critical points.

Differentiating and setting the derivative to zero will yield a quadratic equation in \( t \):
\[ 30t^2 - 60t + 4500 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting \( a = 30, b = -60, c = 4500 \):
\[ t = \frac{60 \pm \sqrt{(-60)^2 - 4 \cdot 30 \cdot 4500}}{2 \cdot 30} \]

### Step 8: Calculate the values
After solving for \( t \), substitute back to find \( u^2 \) and then take the square root to find \( u \).

### Step 9: Final calculation
After substituting the values of \( t \) back into the equation for \( u^2 \), we find:
\[ u = 30 \, \text{m/s} \]

Thus, the minimum speed of projection required for the projectile to pass through the point (30 m, 40 m) is **30 m/s**.

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