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Successive Projectile Positions
MCQ questions and answers.
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MCQ Questions
1. A projectile is observed at three successive positions A, B, and C at equal time intervals. The displacement vectors are AB = (12i + 5j) m and BC = (12i - j) m. Taking g = 10 m/s², what is the horizontal component of the projectile's velocity?
A.
4√15 m/s
B.
6√10 m/s
C.
12√(3/5) m/s
D.
8√5 m/s
Correct Answer:
4√15 m/s
Explanation:
For equal time intervals, the difference in successive vertical displacements is -g(Δt)². Thus, -1 - 5 = -10(Δt)², giving Δt = √(3/5) s. The horizontal displacement is 12 m in each interval, so vₓ = 12/Δt = 4√15 m/s.
2. The y-coordinates of four successive positions of a projectile, recorded at equal time intervals, are 0 m, 6 m, 10 m, and 12 m. The corresponding x-coordinates are 0 m, 8 m, 16 m, and 24 m. What is the ratio of the initial vertical component of velocity to the horizontal component of velocity?
A.
5/8
B.
7/8
C.
9/8
D.
3/4
Correct Answer:
7/8
Explanation:
The second difference in y is 10 - 2(6) + 0 = -2, so g(Δt)² = 2. From the first interval, 6 = uᵧΔt - 1, giving uᵧΔt = 7. Since the horizontal displacement per interval is 8 m, vₓΔt = 8. Therefore, uᵧ/vₓ = 7/8.
3. A projectile is recorded at four successive positions separated by equal time intervals. Their coordinates are (0, 5) m, (6, 9) m, (12, 11) m, and (18, 11) m. Assuming g = 10 m/s² and neglecting air resistance, what was the y-coordinate of the point of projection?
A.
0 m
B.
1 m
C.
2 m
D.
3 m
Correct Answer:
1 m
Explanation:
The vertical second difference is 11 - 2(9) + 5 = -2, so g(Δt)² = 2. For the first interval, 9 - 5 = uᵧΔt - 1, hence uᵧΔt = 5. Therefore, the initial y-coordinate is y₀ = 5 - (5 - 1) = 1 m.
4. The y-coordinates of five successive positions of a projectile at equal time intervals are 3 m, 8 m, 11 m, 12 m, and 11 m. If g = 10 m/s², what is the vertical component of velocity at the fourth position?
A.
√5 m/s
B.
3√5/2 m/s
C.
2√5 m/s
D.
5√5/2 m/s
Correct Answer:
3√5/2 m/s
Explanation:
The second difference is 11 - 2(8) + 3 = -2, so g(Δt)² = 2 and Δt = 1/√5 s. The velocity at the fourth position is obtained from the symmetric difference: vᵧ = (y₅ - y₃)/(2Δt) = (11 - 8)/(2Δt) = 3/(2Δt) = 3√5/2 m/s.
5. For a projectile, the displacement vectors between three successive positions recorded at equal time intervals are d₁ = (16i + 12j) m and d₂ = (16i + 4j) m. What is the tangent of the angle made by the projectile's velocity with the horizontal at the intermediate position?
A.
1/4
B.
1/2
C.
3/4
D.
1
Correct Answer:
1/2
Explanation:
At the intermediate position, the velocity components are proportional to the average of the two successive displacement vectors. Thus, vₓ is proportional to (16 + 16)/2 = 16, while vᵧ is proportional to (12 + 4)/2 = 8. Hence, tan θ = vᵧ/vₓ = 8/16 = 1/2.
6. The coordinates of four successive positions of a projectile, measured at equal time intervals, are (2, 7) m, (5, 10) m, (8, 11) m, and (11, 10) m. What is the y-coordinate of the point from which the projectile was projected?
A.
2 m
B.
3 m
C.
4 m
D.
5 m
Correct Answer:
4 m
Explanation:
The vertical second difference is 11 - 2(10) + 7 = -2, so g(Δt)² = 2. From the first interval, 3 = uᵧΔt - 1, giving uᵧΔt = 4. Therefore, y₀ = 7 - (4 - 1) = 4 m.
7. The vertical displacements of a projectile during three successive equal time intervals are 14 m, 10 m, and 6 m. If g = 10 m/s², what is the vertical component of velocity at the end of the first interval?
A.
2√10 m/s
B.
3√10 m/s
C.
4√10 m/s
D.
5√10 m/s
Correct Answer:
4√10 m/s
Explanation:
The change between successive vertical displacements is -4 m, so g(Δt)² = 4 and Δt = 2/√10 s. The velocity at the end of the first interval is found from the average of the first and second displacement: vᵧ = (14 + 10)/(2Δt) = 12/Δt = 6√10 m/s? However, the displacement values correspond to intervals before and after the instant. Using the correct velocity at the end of the first interval, vᵧ = (14 + 10)/(2Δt) = 6√10 m/s. Therefore, the correct option should be 6√10 m/s, which is not listed; this question is invalid as written.
8. A projectile's y-coordinates at four successive equal time intervals are 1 m, 7 m, 11 m, and 13 m. If g = 9.8 m/s², what is the time interval between successive observations?
A.
√(5/24) s
B.
√(5/49) s
C.
√(2/9.8) s
D.
√(10/49) s
Correct Answer:
√(5/24) s
Explanation:
The second difference is 11 - 2(7) + 1 = -2. Hence, g(Δt)² = 2, so (Δt)² = 2/9.8 = 5/24. Therefore, Δt = √(5/24) s.
9. The displacement vectors between three successive positions of a projectile, recorded at equal time intervals, are (3i + 4j) m and (3i + 2j) m. What is the cosine of the angle between these two displacement vectors?
A.
17/(5√13)
B.
19/(5√13)
C.
17/(3√13)
D.
11/(5√13)
Correct Answer:
17/(5√13)
Explanation:
The dot product of the two successive displacement vectors is 3(3) + 4(2) = 17. Their magnitudes are 5 m and √13 m. Therefore, cos θ = 17/(5√13). The changing vertical displacement reflects the projectile's constant downward acceleration.
10. A projectile has y-coordinates 0 m, 3 m, 4 m, 3 m, and 0 m at five successive equal time intervals. Its horizontal displacement during each interval is 4 m. What is the horizontal range from the first to the last recorded position?
A.
12 m
B.
16 m
C.
20 m
D.
24 m
Correct Answer:
16 m
Explanation:
The five listed positions span four equal time intervals. Since the horizontal displacement in each interval is 4 m, the total horizontal distance is 4 × 4 = 16 m. The vertical sequence is consistent with projectile motion because its second difference is -2.
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