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Standard XII

Physics

Projectile Time, Height and Range

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 = Ax – 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 = xtan $θ$ - $\frac{g x^{2}}{2 u^{2} {c o s}^{2} θ}$
Comparing with given equation
A= tan $θ$ and 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
( $θ$ = $45^{\circ}$ , 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 = 10 ms−2)

The correct option is D 40 : 1Standard equation of projectile motion y = xtanθ - gx22u2cos2θ Comparing with given equation A= tanθ and B = g2u2cos2θ So AB = tanθ×2u2cos2θg = 40 (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 = Ax – B $x^{2}$ where h is height, x is horizontal distance, A and B are constants. The ratio A : B is (g = 10 m $s^{- 2}$ )