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

Question

Quantrex Academy · Accepted Answer

Correct answer: 30

In projectile motion, the equation of the trajectory is

y = xtan \theta - \dfrac{g{x}^{2}}{2{u}^{2}\cos^{2} \theta}

= xtan \theta - \dfrac{g{x}^{2}}{2{u}^{2}}\left(1 + \tan^{2} \theta\right)

Substituting the given values, we have

40 = 30tan \theta - \dfrac{10 \times {\left(30\right)}^{2}}{2{u}^{2}}\left(1 + \tan^{2} \theta\right)

or

900\tan^{2} \theta - \left(6{u}^{2}tan \theta\right) + \left(900 + 8{u}^{2}\right) = 0

The value of

tan \theta

will be real if

{\left(6{u}^{2}\right)}^{2} \ge 4 \times 900 \times \left(900 + 8{u}^{2}\right)

or

{u}^{4} \ge 100\left(900 + 8{u}^{2}\right)

or

{u}^{4} - 800{u}^{2} \ge 90000

or

{\left({u}^{2} - 400\right)}^{2} - 160000 \ge 90000

or

{\left({u}^{2} - 400\right)}^{2} \ge 250000

{u}^{2} - 400 \ge 500 ⇒ {u}^{2} \ge 900

⇒ u \ge 30

{m s}^{- 1}

A body is projected from the ground with some speed at some angle...

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