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 axes respectively and the point of projection as origi

Question

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 axes respectively and the point of projection as origin, calculate the minimum speed (in m s - 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 = 10 m s - 2

TestHub · Accepted Answer

In projectile motion, the equation of the trajectory is

$y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

$= x \tan θ - \frac{g x^{2}}{2 u^{2}} (1 + \tan^{2} θ)$

Substituting the given values, we have

$40 = 30 \tan θ - \frac{10 \times {(30)}^{2}}{2 u^{2}} (1 + \tan^{2} θ)$

or $900 \tan^{2} θ - (6 u^{2} \tan θ) + (900 + 8 u^{2}) = 0$

The value of $\tan θ$ will be real if

${(6 u^{2})}^{2} \geq 4 \times 900 \times (900 + 8 u^{2})$

or $u^{4} \geq 100 (900 + 8 u^{2})$

or $u^{4} - 800 u^{2} \geq 90000$

or ${(u^{2} - 400)}^{2} - 160000 \geq 90000$

or ${(u^{2} - 400)}^{2} \geq 250000$

$u^{2} - 400 \geq 500 \Rightarrow u^{2} \geq 900$

$\Rightarrow u \geq 30$ ${m s}^{- 1}$

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