The trajectory of a projectile in a vertical plane is $ y=a x-b x^{2} $, where $ a, b $ are constants, and $ x $ and $ y $ are respectively the horizontal and vertical distances of the projectile from the point of projection. Find the maximum height $ H_{\max } $ attained by the projectile and the angle of projection from the horizontal.

Question

The trajectory of a projectile in a vertical plane is $ y=a x-b x^{2} $, where $ a, b $ are constants, and $ x $ and $ y $ are respectively the horizontal and vertical distances of the projectile from the point of projection. Find the maximum height $ H_{\max } $ attained by the projectile and the angle of projection from the horizontal., $ H_{\max }=\frac{a^{2}}{4 b} $ $ =\tan ^{-1} a $, $H_{\max } =\frac{a}{2 b} $ $ =\tan ^{-1} b $, $ H_{\max }=\frac{a^{2}}{2 b} $ $ \theta=\tan ^{-1}\left(\sqrt{a^{2}+1}\right) $, $ H_{\max }=\frac{a}{4 b^{2}} $ $ \theta=\tan ^{-1} a $

MARKS App · Accepted Answer

Trajectory of the projectile in a vertical plane - y = a x - b x 2 Comparing it with equation of a projectile y = x tan θ - g x 2 2 u 2 cos 2 θ ⇒ tan θ = a ....(i) ⇒ θ = tan - 1 ( a ) and g 2 u 2 cos 2 θ = b .....(ii) From (ii) g 2 b 2 cos 2 θ = g a 2 + 1 2 b ....(iii) ∴ cos θ = 1 a 2 + 1 Maximum height attained, H max = u 2 sin 2 θ 2 a .....(iv) From (ii) g 2 b = u 2 cos 2 θ ⇒ g 2 b = u 2 - u 2 sin 2 θ ⇒ u 2 sin 2 θ = g a 2 + 1 2 b - g 2 b = g a 2 2 b ......(v) From (iv) and (v) H max = g a 2 2 b × 2 g = a 2 4 b Alternatively d y d x = a - 2 b x = 0 (For Maximum Height) ⇒ x = a 2 h Substituting the value of x in y = a x - b x 2 to find max height H max = a a 2 b - b a 2 4 b 2 = a 2 2 b - a 2 4 b = a 2 4 b

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