SATHEE: Kinematics Question 611

Kinematics Question 611

Question: The equation of trajectory of projectile is given by $y = \frac{x}{\sqrt{3}} - \frac{g}{x^{2}} 20$ , where x and y are in meter. The maximum range of the projectile is

Options:

A) $\frac{8}{3} m$

B) $\frac{4}{3} m$

C) $\frac{3}{4} m$

D) $\frac{3}{8} m$

Show Answer

Answer:

Correct Answer: B

Solution:

[b] Comparing the given equation with the equation of trajectory of a projectile, $y = x \tan θ - \frac{g}{x^{2}} 2 u^{2} \cos^{2} θ,$ we get, $tan θ = \frac{1}{\sqrt{3}} \Rightarrow θ = 30^{\circ}$ and $2 u^{2} \cos^{2} θ = 20 \Rightarrow u^{2} = \frac{20}{2 \cos^{2} θ} = \frac{40}{3}$ Now, $R_{max} = \frac{u_{2}}{g} = \frac{40}{3 \times 10} = \frac{4}{3} m$

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Kinematics Question 611

Question: The equation of trajectory of projectile is given by y=x3−gx220 , where x and y are in meter. The maximum range of the projectile is

Options:

Answer:

Solution:

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Question: The equation of trajectory of projectile is given by $y = \frac{x}{\sqrt{3}} - \frac{g}{x^{2}} 20$ , where x and y are in meter. The maximum range of the projectile is