Q.Three balls projected upwards with the same initial speed at angle θ = 30°, 45° and 60° respectively. Let the range of the ball be expressed by R₀. Then:

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 …