Q.Two objects are projected at angles 30 degrees and 60 degrees respectively with respect to the horizontal direction. The range of two objects are denoted as R30 and R60. Choose the correct relation from the following :

Complementary projection angles (theta and 90 - theta) produce equal horizontal ranges for the same launch speed, since sin(2theta) is the same for both.

Horizontal range of a projectile: R = (v^2 sin(2theta)) / g

For body A, theta_A = 30 degrees: sin(2 x 30) = sin(60) = sqrt(3)/2

For body B, theta_B = 60 degrees: sin(2 x 60) = sin(120) = sqrt(3)/2

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The maximum horizontal range of a projectile is R_max = u²/g, occurring at a launch angle of 45°.

For a projectile launched with initial speed u at angle θ to the horizontal, the horizontal range is R = (u² sin 2θ)/g. Since sin 2θ has a maximum value of 1 (when 2θ = 90°, i.e. θ = 45°), the range is maximum when θ = 45°, giving R_max = u²/g. This is why, for a given launch …

Complementary projection angles θ and (90°−θ) produce the same horizontal range, though the maximum height and time of flight differ.

For a projectile launched with speed u at angle θ to the horizontal, the horizontal range is:

R = (u² sin 2θ) / g

For the complementary angle (90°−θ):

R' = (u² sin[2(90°−θ)]) / g = (u² sin(180°−2θ)) / g

Using sin(180°−x) = sin x:

R' = (u² sin 2θ) / g = R

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The range of a projectile, $R = \dfrac{u^2\sin 2\theta}{g}$, is maximum when the angle of projection is 45°.

For a projectile launched with speed $u$ at angle $\theta$ to the horizontal, the horizontal range is

$R = \dfrac{u^2 \sin 2\theta}{g}$

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Range R = (u² sin2θ)/g is the same for θ and 90° − θ because sin2θ = sin(2(90°−θ)) = sin(180°−22θ) has the same value.

The horizontal range of a projectile launched with speed u at angle θ is:

$R = \frac{u^2 \sin 2\theta}{g}$

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Two projection angles give the same horizontal range whenever they are complementary (add up to 90 deg); only 20 deg and 70 deg satisfy this among the options.

The horizontal range of a projectile is $R = \dfrac{u^2 \sin(2\theta)}{g}$.

For two angles $\theta$ and $(90^\circ - \theta)$: $\sin(2\theta) = \sin(2(90^\circ-\theta)) = \sin(180^\circ - 2\theta) = \sin(2\theta)$, so the range is identical for both.

Checking the options: (A) 20+60=80 deg (not complementary), …

True. The horizontal range R = u^2 sin(2theta)/g is the same for complementary angles theta and (90-theta) because sin(2theta) = sin(180-2theta) = sin(2(90-theta)).

For a projectile launched with speed u at angle theta to the horizontal, the horizontal range is:

R = (u^2 sin(2 theta)) / g

For the complementary angle (90 degrees - theta):

R' = (u^2 sin(2(90-theta))) / g = (u^2 sin(180 - 2theta)) / g = (u^2 sin(2theta)) / g = R …

Complementary angles of projection (angles that add up to 90 degrees) give equal horizontal range; only the pair (30 deg, 60 deg) among the given options is complementary.

The horizontal range of a projectile is R = u^2 sin(2theta)/g. This depends on theta only through sin(2theta). For two angles theta1 and theta2 = 90 deg - theta1:

2theta2 = 180 deg - 2theta1, so sin(2theta2) = sin(180 deg - 2theta1) = sin(2*theta1)

Hence R is the same for theta1 and 90 deg - theta1.

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Maximum horizontal range occurs at a projection angle of 45°.

Derivation. The horizontal range of a projectile launched with speed $u$ at angle $\theta$ to the horizontal is:

$R = \frac{u^2\sin2\theta}{g}$ …

Range of a projectile is R = u²sin(2θ)/g. For projection angles θ and (90°-θ), the value of sin(2θ) is the same, so their ranges are equal — 30° and 60° are exactly such a complementary pair.

Range formula for a projectile launched with speed u at angle θ to the horizontal (same launch and landing height):

R = u² sin(2θ) / g

For θ = 30°: R30 = u² sin(60°) / g

For θ = 60°: R60 = u² sin(120°) / g

But sin(120°) = sin(180° − 120°) = sin(60°), so sin(60°) = sin(120°).

Therefore R30 = u² sin(60°)/g = u² sin(120°)/g = R60.

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Range $R = \dfrac{u^2\sin 2\theta}{g}$ gives the same value for $\theta$ and $90^\circ - \theta$, since $\sin 2\theta = \sin(180^\circ - 2\theta)$.

The horizontal range of a projectile launched with speed $u$ at angle $\theta$ is:

$R = \frac{u^2 \sin 2\theta}{g}$

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Range R = u²sin(2θ)/g is maximum at θ=45°; complementary angles (30° & 60°) give equal, smaller ranges. So R45 > R30 = R60.

For a projectile launched with speed u at angle θ to the horizontal, the horizontal range is

$R = \dfrac{u^2\sin(2\theta)}{g}$

Compute sin(2θ) for each angle:

• θ = 30° → sin(60°) = √3/2 ≈ 0.866
• θ = 45° → sin(90°) = 1
• θ = 60° → sin(120°) = √3/2 ≈ 0.866 …