The parabolic path of projectile is Y - x -3- x 2-60 its maximum height is A- 5 mB- 10 mC- 15 mD- 20 m

Comparing with the standard equation of trajectory of a projectile tan θ = 1/√(3);θ = 30^0 R = a/b = 60/√(3) H = Rtan θ/4=(60/√(3))(1/√(3))/4= 5 m ...

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Standard XII

Physics

Introduction to Projectile Motion

The parabolic...

Question

The parabolic path of projectile is Y= $\frac{x}{\sqrt{3}} - \frac{x^{2}}{60}$ its maximum height is

5 m

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10 m

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15 m

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20 m

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Solution

The correct option is A 5 m
Comparing with the standard equation of trajectory of a projectile
$t a n θ = \frac{1}{\sqrt{3}}; θ = 30^{0} R = \frac{a}{b} = \frac{60}{\sqrt{3}} H = \frac{R t a n θ}{4} = \frac{(\frac{60}{\sqrt{3})} (\frac{1}{\sqrt{3}})}{4} = 5 m$

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The parabolic path of projectile is Y= x√3−x260 its maximum height is

The correct option is A 5 mComparing with the standard equation of trajectory of a projectile tan θ=1√3;θ=300R=ab=60√3H=Rtan θ4=(60√3)(1√3)4=5 m

The parabolic path of projectile is Y= $\frac{x}{\sqrt{3}} - \frac{x^{2}}{60}$ its maximum height is

The correct option is A 5 m
Comparing with the standard equation of trajectory of a projectile
$t a n θ = \frac{1}{\sqrt{3}}; θ = 30^{0} R = \frac{a}{b} = \frac{60}{\sqrt{3}} H = \frac{R t a n θ}{4} = \frac{(\frac{60}{\sqrt{3})} (\frac{1}{\sqrt{3}})}{4} = 5 m$