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A body is projected from the ground with...

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)`

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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: \[ \boxed{30 \text{ m/s}} \]
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