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 required for a body to pass through the point (30 m, 40 m), we can follow these steps:

### Step 1: Understand the equations of projectile motion
The equation of the trajectory of a projectile is given by:
\[ y = x \tan(\theta) - \frac{g}{2} \frac{x^2}{u^2 \cos^2(\theta)} \]
where:
- \(y\) is the vertical position,
- \(x\) is the horizontal position,
- \(g\) is the acceleration due to gravity (10 m/s² in this case),
- \(u\) is the initial speed of projection,
- \(\theta\) is the angle of projection.

### Step 2: Substitute the known values
We need to find the minimum speed \(u\) such that the projectile passes through the point (30 m, 40 m). Thus, we substitute \(x = 30\) m and \(y = 40\) m into the trajectory equation:
\[ 40 = 30 \tan(\theta) - \frac{10}{2} \frac{30^2}{u^2 \cos^2(\theta)} \]
This simplifies to:
\[ 40 = 30 \tan(\theta) - \frac{1500}{u^2 \cos^2(\theta)} \]

### Step 3: Express \(\tan(\theta)\) in terms of \(p\)
Let \(p = \tan(\theta)\). Then, we can express \(\cos^2(\theta)\) using the identity:
\[ \cos^2(\theta) = \frac{1}{1 + p^2} \]
Substituting this into the equation gives:
\[ 40 = 30p - \frac{1500(1 + p^2)}{u^2} \]

### Step 4: Rearranging the equation
Rearranging the equation to isolate \(u^2\):
\[ 40u^2 = 30pu - 1500(1 + p^2) \]
\[ u^2 = \frac{30p - 40}{1500} \cdot (1 + p^2) \]

### Step 5: Differentiate to find the minimum speed
To find the minimum speed, we need to minimize \(u^2\) with respect to \(p\). We differentiate \(u^2\) with respect to \(p\) and set the derivative to zero:
\[ \frac{d(u^2)}{dp} = 0 \]
This leads to:
\[ \frac{d}{dp}\left(\frac{30p - 40}{1500}(1 + p^2)\right) = 0 \]

### Step 6: Solve the derivative equation
After differentiating and simplifying, we arrive at a quadratic equation in \(p\):
\[ 3p^2 - 8p - 3 = 0 \]
Using the quadratic formula \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ p = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ p = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm 10}{6} \]
This gives us two possible values for \(p\):
\[ p = 3 \quad \text{or} \quad p = -\frac{1}{3} \]

### Step 7: Finding the minimum speed
We take \(p = 3\) (since \(p\) must be positive) and substitute it back into the equation for \(u^2\):
\[ u^2 = \frac{450(1 + 3^2)}{3 \cdot 3 - 4} = \frac{450 \cdot 10}{5} = 900 \]
Thus, the minimum speed \(u\) is:
\[ u = \sqrt{900} = 30 \text{ m/s} \]

### Final Answer
The minimum speed of projection required for the body to pass through the point (30 m, 40 m) is:
\[ \boxed{30 \text{ m/s}} \]