Solution: The plane's velocity relative to air is 240 km/h due north. The wind blows from west to east at 100 km/h — that's a velocity of 100 km/h pointing east. Ground velocity = plane's velocity (air) + wind velocity, added as vectors (north and east components): Magnitude = √(240² + 100²) = √(57600 + 10000) = √67600 = 260 km/h Direction: angle east of north = arctan(100/240) = arctan(0.417) ≈ 23° east of north. So the ground velocity is 260 km/h, 23° E of N.

Solution: At the highest point of a projectile's path, the velocity is purely horizontal (the vertical component has momentarily dropped to zero), while the acceleration due to gravity is always purely vertical (downward). A horizontal vector and a vertical vector are perpendicular — the angle between them is 90°.

Solution: this question notes "(more than one correct)", and indeed two of the four statements are true (the second choice even has the word "correct" accidentally left inside its own text in the source data, alongside the one flagged correct here): • With respect to an observer on the GROUND: the package keeps the plane's horizontal velocity at the moment of release, and also falls under gravity — combining constant horizontal motion with vertically accelerated motion traces a parabola, exactly like ordinary projectile motion. • With respect to an observer travelling IN THE PLANE: since the plane and the package share the same horizontal velocity at release, and the plane continues at that same horizontal velocity while the package doesn't accelerate horizontally at all, their horizontal separation stays zero — to the pilot, the package appears to fall straight down, a vertical straight line. So: parabolic path relative to the ground, and a vertical straight-line path relative to an observer in the plane, are both correct.

Solution: Height h is reached twice: once going up (at time a) and once coming down (at time b), both measured from the moment of launch. Both times satisfy h = ut − ½gt², so a and b are the two roots of the quadratic gt² − 2ut + 2h = 0. For a quadratic gt² − 2ut + 2h = 0, the sum of the roots is (2u)/g, so: a + b = 2u/g → u = g(a+b)/2.

Solution: Let both vectors have magnitude A. Using |A+B|² = A² + B² + 2AB cos θ with A = B: |Resultant|² = A² + A² + 2A²cos θ = 2A²(1 + cos θ) Given the resultant magnitude equals A itself: 2A²(1+cos θ) = A² → 2(1+cos θ) = 1 → cos θ = −0.5 → θ = 120°.

Solution: Both statements are true, and the Reason directly explains the Assertion: near the Earth's surface, gravity is essentially constant (uniform), which is exactly the condition under which projectile motion works out to a parabola. At very great heights, gravity weakens with distance (varies as 1/r², following the inverse-square law), and under such a varying central force the trajectory becomes elliptical instead — the same physics that governs orbital motion (Kepler's laws). So the changing nature of the gravitational force at great heights is precisely why the path becomes elliptical, making the Reason the correct explanation of the Assertion.

Solution: The woman's speed relative to the ground is her walking speed plus the escalator's speed, since both act in the same direction: 1.5 + 1.0 = 2.5 m/s. Time = distance / speed = 50.0 m / 2.5 m/s = 20 s.

Solution: For the mass to move in a horizontal circle, the net force on it must point horizontally, toward the center of that circle (centripetal force). The only forces on the mass are gravity (straight down) and the string's tension (along the string). If the string stayed perfectly vertical, tension would be purely vertical too, leaving nothing to supply the needed horizontal centripetal force — motion in a circle would be impossible. So the string must tilt away from the vertical, giving tension both a vertical component (balancing gravity) and a horizontal component (supplying the centripetal force) — this is the classic "conical pendulum" setup.

Solution: Both particles start with the same initial VERTICAL velocity (zero) and experience the same vertical acceleration (g) throughout the fall — B's horizontal velocity affects only its horizontal position, not its vertical motion at all (horizontal and vertical motions are independent). Since their vertical motions are identical, both particles cover the same vertical drop in the same amount of time — they hit the ground simultaneously.

Solution: Let the projectile be launched with speed v at angle θ. Time to reach the highest point: t = v sin θ / g. Horizontal displacement by then: x = v cos θ · t = v² sin θ cos θ / g Vertical displacement (max height): y = v² sin²θ / (2g) Displacement magnitude = √(x² + y²) = (v² sin θ / g)·√(cos²θ + sin²θ/4) Average velocity = displacement / time = v·√(cos²θ + sin²θ/4) Using sin²θ = 1 − cos²θ and simplifying: = v·√[(4cos²θ + 1 − cos²θ)/4] = (v/2)·√(1 + 3cos²θ)

Solution: 60° and 30° are complementary angles (they add to 90°). For the same launch speed v: Range R = v²sin(2θ)/g — for θ = 60°, sin(2θ) = sin120° = √3/2; for θ = 30°, sin(2θ) = sin60° = √3/2. Same value, so the RANGE is identical for both. Time of flight T = 2v sinθ/g and Maximum height H = v²sin²θ/(2g) both depend on sinθ directly, which is different for 60° (√3/2) and 30° (1/2) — so T and H differ between the two. Only the range stays the same.

Solution: With air resistance neglected, the only force acting throughout the flight is gravity, which is constant in both magnitude and direction (always straight down). Since there's no horizontal force at all, the horizontal component of velocity never changes. So the two quantities that stay constant throughout the flight are the horizontal component of velocity, and the acceleration itself (magnitude g, direction always downward).

Solution: For a particle in circular motion, the centripetal acceleration is a = v²/r. For a given (fixed) speed v, this shows a is inversely proportional to the radius r — a smaller circle requires a larger centripetal acceleration to keep the particle turning at the same speed.

Solution: For a projectile launched at speed v with the same range R achievable at two complementary angles θ and (90°−θ), the times of flight are: t₁ = 2v sinθ/g and t₂ = 2v cosθ/g (since sin(90°−θ) = cosθ) Product: t₁t₂ = 4v² sinθ cosθ/g² = 2v² sin(2θ)/g² Since Range R = v² sin(2θ)/g, we have v² sin(2θ) = Rg, so: t₁t₂ = 2(Rg)/g² = 2R/g So t₁t₂ is directly proportional to R.

Solution: Position vector: r(t) = [4 + 2t²] î + 5t ĵ (metres, t in seconds). At t = 0: r(0) = 4î + 0ĵ At t = 2 s: r(2) = [4 + 2(2)²]î + 5(2)ĵ = 12î + 10ĵ Displacement: Δr = 8î + 10ĵ Average velocity = Δr / Δt = 4î + 5ĵ Magnitude = √(4² + 5²) = √41 ≈ 6.4 m/s, at arctan(5/4) ≈ 51° from the x-axis. (Note: with the coefficients exactly as given in the question — 2 m/s² and 5 m/s — the correctly computed answer is about 6.4 m/s at ≈51°, which doesn't cleanly match any of the printed options including the one marked correct here; this looks like a coefficient got altered somewhere in the source data's scraping/transcription. The method above is the standard, correct approach for this kind of average-velocity-from-position-vector problem regardless.)

Solution: Set maximum height equal to range: v²sin²θ/(2g) = v²sin(2θ)/g = 2v² sinθ cosθ/g sin²θ/2 = 2 sinθ cosθ Dividing both sides by sinθ (θ ≠ 0): sinθ/2 = 2cosθ → tanθ = 4 θ = arctan(4) ≈ 76°.

Solution: Squaring |A+B| = |A|: A² + B² + 2AB cosθ = A² → B² + 2AB cosθ = 0 → B = −2A cosθ (for B ≠ 0). This requires cosθ to be negative — i.e. the angle between A and B must be obtuse (greater than 90°), with the specific magnitude relation |B| = 2|A|cos(180°−θ). Among the listed choices, "A, B are antiparallel" (θ = 180°, cosθ = −1, giving |B| = 2|A|) is the qualitatively closest description of this obtuse-angle, opposing-direction relationship, though strictly the exact condition depends on the magnitude ratio between A and B, not just their being exactly antiparallel.

Solution: For a particle moving in a circle at constant angular speed (uniform circular motion), the acceleration is centripetal — it points radially inward, toward the center of the circle, which is NORMAL (perpendicular) to the circle, not tangent to it. So "the acceleration vector is tangent to the circle" is the false statement; the other three (normal to the circle, pointing to the center, and perpendicular to the velocity) are all true descriptions of centripetal acceleration.