physics : class11 Maximum Range Condition : Projectile Motion : Kinematics MCQ Questions and Answers
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Maximum Range Condition
MCQ questions and answers.
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MCQ Questions
1. A projectile is launched from level ground with a fixed speed u and lands at the same level. At what launch angle is its horizontal range maximum?
A.
30°
B.
45°
C.
60°
D.
90°
Correct Answer:
45°
Explanation:
For a projectile landing at its launch level, the range is R = (u² sin 2θ)/g. Since sin 2θ is maximum when 2θ = 90°, the maximum range occurs at θ = 45°.
2. A projectile is launched with speed 30 m/s from level ground. Taking g = 10 m/s², what is its maximum possible horizontal range?
A.
45 m
B.
60 m
C.
90 m
D.
120 m
Correct Answer:
90 m
Explanation:
The maximum range is Rmax = u²/g because sin 2θ = 1 at θ = 45°. Thus Rmax = 30²/10 = 90 m.
3. For a projectile launched and landing at the same level, the range obtained at an acute angle θ is half its maximum possible range. If θ is the smaller of the two possible angles, what is θ?
A.
15°
B.
30°
C.
45°
D.
75°
Correct Answer:
15°
Explanation:
Since R/Rmax = sin 2θ, the condition R = Rmax/2 gives sin 2θ = 1/2. The smaller acute solution is 2θ = 30°, so θ = 15°.
4. A projectile has a maximum horizontal range of 180 m when launched from level ground. Taking g = 10 m/s², what launch speed is required?
A.
30 m/s
B.
30√2 m/s
C.
60 m/s
D.
90 m/s
Correct Answer:
30√2 m/s
Explanation:
At the maximum-range condition, Rmax = u²/g. Therefore u = √(gRmax) = √(10 × 180) = √1800 = 30√2 m/s.
5. A projectile is launched with speed 20 m/s at the angle for maximum range. Taking g = 10 m/s², what is the maximum height reached?
A.
5 m
B.
10 m
C.
15 m
D.
20 m
Correct Answer:
10 m
Explanation:
Maximum range occurs at 45°, so the initial vertical component is u/√2. Hence H = (u² sin²45°)/(2g) = (400 × 1/2)/(20) = 10 m.
6. A projectile launched from level ground at 30° has a horizontal range of 50 m. What would be the maximum range possible with the same launch speed?
A.
50√3 m
B.
100 m
C.
100/√3 m
D.
200/√3 m
Correct Answer:
100/√3 m
Explanation:
For 30°, R = Rmax sin 60° = Rmax(√3/2). Thus Rmax = 50 ÷ (√3/2) = 100/√3 m.
7. For a projectile launched and landing at the same level, the horizontal and vertical components of its initial velocity are equal. Which conclusion follows?
A.
The projectile has zero range
B.
The projectile is launched for maximum range
C.
The projectile reaches maximum height at launch
D.
The projectile has maximum range at 30°
Correct Answer:
The projectile is launched for maximum range
Explanation:
Equal horizontal and vertical components imply tan θ = 1, so θ = 45°. For a fixed launch speed and equal launch and landing levels, 45° is the maximum-range condition.
8. Two projectiles are launched with the same speed from level ground at 20° and 70°, respectively. How do their ranges compare with the maximum range possible for that speed?
A.
Each range equals the maximum range
B.
Each range is sin 20° times the maximum range
C.
Each range is sin 40° times the maximum range
D.
Their ranges are different because only 45° gives any range
Correct Answer:
Each range is sin 40° times the maximum range
Explanation:
For either angle, R/Rmax = sin 2θ. At 20°, this is sin 40°, while at 70° it is sin 140° = sin 40°. Thus both ranges are equal and equal to sin 40° of the maximum.
9. A projectile has a maximum range of 120 m. What range does it achieve when launched at 30° instead of the maximum-range angle?
A.
30√3 m
B.
60 m
C.
60√3 m
D.
120√3 m
Correct Answer:
60√3 m
Explanation:
The range at angle θ is R = Rmax sin 2θ. For θ = 30°, R = 120 sin 60° = 120(√3/2) = 60√3 m.
10. A projectile launched from level ground has a range equal to 3/4 of its maximum possible range. What is the smaller launch angle that produces this range?
A.
18.4°
B.
24.3°
C.
37.5°
D.
65.7°
Correct Answer:
24.3°
Explanation:
The condition R = 3Rmax/4 gives sin 2θ = 3/4. The smaller solution is 2θ = sin⁻¹(3/4) ≈ 48.6°, hence θ ≈ 24.3°.
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