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Standard XII

Physics

The Equation for the Path of Projectile

An object is ...

Question

An object is projected with a velocity of 20 m/s making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory is $h = A x - B x^{2}$ where h is height, x is horizontal distance, A and B are constants. The ratio A : B is $(g = 10 m s^{- 2})$

1 : 5

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5 : 1

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1 : 40

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40 : 1

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Solution

The correct option is D 40 : 1
Standard equation of projectile motion
$y = x t a n θ - \frac{g x^{2}}{2 u^{2} c o s^{2} θ}$
Comparing with given equation
$A = t a n θ a n d B = \frac{g}{2 u^{2} c o s^{2} θ}$
So $\frac{A}{B} = \frac{t a n θ \times 2 u^{2} c o s^{2} θ}{g} = 40$
$(A s θ = 45^{\circ}, u = 20 m / s, g = 10 m / s^{2})$

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An object is projected with a velocity of 20 m/s making an angle of 45∘ with horizontal. The equation for the trajectory is h=Ax−Bx2 where h is height, x is horizontal distance, A and B are constants. The ratio A : B is (g=10ms−2)

The correct option is D 40 : 1Standard equation of projectile motion y=xtanθ−gx22u2cos2θ Comparing with given equation A=tan θ and B=g2u2cos2θ So AB=tan θ×2u2cos2θg=40 (As θ=45∘,u=20 m/s,g=10 m/s2)

An object is projected with a velocity of 20 m/s making an angle of $45^{\circ}$ with horizontal. The equation for the trajectory is $h = A x - B x^{2}$ where h is height, x is horizontal distance, A and B are constants. The ratio A : B is $(g = 10 m s^{- 2})$