Question

A mango in a tree is located at a horizontal and vertical distance of

30 m

and

40 m

, respectively, from the point of projection of a stone. Find the minimum speed and the critical angle of projection of the stone so as to hit the mango.

Accepted Answer

Here, the coordinate of $P (30 m, 40 m)$ .

By using equation of path

$y = x \tan (θ) - \frac{g x^{2}}{2 u^{2} \cos^{2} θ} \Rightarrow y = x \tan (θ) - \frac{g x^{2} (s e c^{2} θ)}{2 u^{2}} \Rightarrow y = x \tan (θ) - \frac{g x^{2} (1 + \tan^{2} (θ))}{2 u^{2}} \Rightarrow \tan^{2} (θ) - \frac{2 u^{2}}{g x} \tan (θ) + (1 + \frac{2 u^{2} y}{g x^{2}}) = 0 . . . (1)$

For real roots

$b^{2} - 4 a c \geq 0 \Rightarrow {(\frac{2 u^{2}}{g x})}^{2} - 4 (1 + \frac{2 u^{2} y}{g x^{2}}) \geq 0 ∵ x = 30 & y = 40 \Rightarrow {(\frac{2 u^{2}}{10 \times 30})}^{2} - 4 (1 + \frac{2 u^{2} (40)}{10 {(30)}^{2}}) \geq 0$

The minimum velocity required to hit the mango is

${(\frac{2 u^{2}}{10})}^{2} - 4 (1 + \frac{2 u^{2} (40)}{10 {(30)}^{2}}) = 0 \Rightarrow u^{4} - 800 u^{2} - 90000 = 0 \Rightarrow u^{4} - 900 u^{2} + 100 u^{2} - 90000 = 0 \Rightarrow (u^{2} - 900) (u^{2} + 100) = 0 \Rightarrow u^{2} - 900 \Rightarrow u = 30 m s^{- 1}$

From equation $(1)$

$\tan^{2} (θ) - \frac{2 u^{2}}{g x} \tan (θ) + (1 + \frac{2 u^{2} y}{g x^{2}}) = 0 \Rightarrow \tan^{2} (θ) - \frac{2 {(30)}^{2}}{10 (30)} \tan (θ) + (1 + \frac{2 {(30)}^{2} 40}{10 {(30)}^{2}}) = 0 \Rightarrow \tan^{2} (θ) - 6 \tan (θ) + (9) = 0 \Rightarrow {(\tan (θ) - 3)}^{2} = 0 \Rightarrow \tan (θ) = 3$
Hence, $θ = \tan^{- 1} (3)$

B M Sharma Solutions for Chapter: Kinematics II, Exercise 38: ILLUSTRATION 5.36

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 38: ILLUSTRATION 5.36

Questions from B M Sharma Solutions for Chapter: Kinematics II, Exercise 38: ILLUSTRATION 5.36 with Hints & Solutions

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