physics : class11 Projectile Equation of Trajectory : Projectile Motion : Kinematics MCQ Questions and Answers
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Projectile Equation of Trajectory
MCQ questions and answers.
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MCQ Questions
1. A projectile is launched from the origin with speed u at an angle θ to the horizontal. Its trajectory is observed to be y = x - x²/20, where x and y are in metres. Taking g = 10 m/s², which pair gives u and θ?
A.
10 m/s and 30°
B.
10√2 m/s and 45°
C.
20 m/s and 45°
D.
5√2 m/s and 60°
Correct Answer:
10√2 m/s and 45°
Explanation:
Comparing y = x tanθ − gx²/(2u²cos²θ) with y = x − x²/20 gives tanθ = 1, so θ = 45°. Also, 10/(2u² × 1/2) = 1/20, which gives u² = 200 and hence u = 10√2 m/s.
2. The equation of trajectory of a projectile is y = √3x − x²/4, with x and y measured in metres. If g = 10 m/s², what is the initial speed of projection?
A.
2√5 m/s
B.
4√5 m/s
C.
5√2 m/s
D.
10 m/s
Correct Answer:
4√5 m/s
Explanation:
Comparison with y = x tanθ − gx²/(2u²cos²θ) gives tanθ = √3, so cos²θ = 1/4. The quadratic coefficient gives 10/(2u² × 1/4) = 1/4, leading to u² = 80 and u = 4√5 m/s.
3. A projectile has the trajectory y = 2x − x²/10, where distances are in metres. At what horizontal positions does it reach a height of 5 m?
A.
x = 5 ± √5 m
B.
x = 10 ± 5√2 m
C.
x = 10 ± √50 m
D.
x = 5 ± 5√2 m
Correct Answer:
x = 10 ± 5√2 m
Explanation:
Setting y = 5 gives 5 = 2x − x²/10, or x² − 20x + 50 = 0. Solving this quadratic yields x = 10 ± √50 = 10 ± 5√2 m.
4. The trajectory of a projectile is y = 3x/4 − x²/16. At the point where x = 4 m, what is the slope of the trajectory and what does it imply about the instantaneous direction of motion?
A.
Slope = 1/4; the projectile is rising
B.
Slope = 1/2; the projectile is rising
C.
Slope = 1/4; the projectile is falling
D.
Slope = 1/2; the projectile is falling
Correct Answer:
Slope = 1/4; the projectile is rising
Explanation:
For y = 3x/4 − x²/16, dy/dx = 3/4 − x/8. At x = 4 m, dy/dx = 3/4 − 1/2 = 1/4, which is positive, so the projectile is still rising.
5. Two projectiles are launched from the same point and have trajectory equations y = x − x²/20 and y = √3x − x²/15 respectively. At what nonzero horizontal distance do their trajectories intersect?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
Correct Answer:
15 m
Explanation:
Equating the two equations gives x − x²/20 = √3x − x²/15 only if the linear coefficients are treated numerically; however, the stated equations do not yield a simple listed intersection because √3 is not equal to a rational coefficient. Therefore, this question is invalid as written.
6. A projectile follows the equation y = x − x²/8, with x and y in metres. What are the horizontal range and maximum height of the projectile?
A.
8 m and 1 m
B.
8 m and 2 m
C.
16 m and 2 m
D.
16 m and 4 m
Correct Answer:
8 m and 2 m
Explanation:
The nonzero root of y = x − x²/8 is x = 8 m, giving the range. The vertex occurs at x = −b/(2a) = 4 m; substituting gives y = 4 − 16/8 = 2 m, the maximum height.
7. The trajectory of a projectile is y = 2x − x²/5. At what horizontal distance from the point of projection is the projectile moving horizontally?
A.
2.5 m
B.
5 m
C.
7.5 m
D.
10 m
Correct Answer:
5 m
Explanation:
The trajectory is horizontal where dy/dx = 0. Since dy/dx = 2 − 2x/5, setting this equal to zero gives x = 5 m. This is the horizontal coordinate of the vertex.
8. A projectile has trajectory y = (√3/2)x − x²/12. What fraction of its horizontal range corresponds to the point where its height is maximum?
A.
1/4
B.
1/3
C.
1/2
D.
2/3
Correct Answer:
1/2
Explanation:
For any trajectory of the form y = ax − bx², the vertex occurs at x = a/(2b), while the nonzero intercept is R = a/b. Therefore, the maximum height occurs at x = R/2, so the required fraction is 1/2.
9. A projectile is launched from the point (0, 2 m), and its trajectory is y = 2 + x − x²/20, where x and y are measured in metres. What is its horizontal range before it reaches the ground?
A.
10 m
B.
12 m
C.
20 m
D.
22 m
Correct Answer:
20 m
Explanation:
The projectile reaches the ground when y = 0. Thus, 0 = 2 + x − x²/20, giving x² − 20x − 40 = 0. The positive root is x = 10 + 2√35 m, not 20 m. Therefore, this question is invalid as written.
10. Two projectiles have trajectory equations y = x − x²/20 and y = 2x − x²/10 respectively. Which statement correctly compares their trajectories?
A.
They have the same range but different maximum heights
B.
They have the same maximum height but different ranges
C.
They have both the same range and the same maximum height
D.
Their ranges and maximum heights are both different
Correct Answer:
They have both the same range and the same maximum height
Explanation:
For the first trajectory, the nonzero intercept is x = 20 and the vertex height is 1²/(4 × 1/20) = 5 m. For the second, the nonzero intercept is x = 20 and the vertex height is 2²/(4 × 1/10) = 10 m. Hence they have the same range but different maximum heights, so the correct option should be optionA; the supplied correct_answer is inconsistent.
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