physics : class11 Maximum Height : Projectile Motion : Kinematics MCQ Questions and Answers
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Maximum Height
MCQ questions and answers.
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MCQ Questions
1. For a projectile launched with initial vertical velocity component u_y, what is its maximum height above the point of projection?
A.
u_y^2/g
B.
u_y^2/(2g)
C.
2u_y^2/g
D.
u_y/g
Correct Answer:
u_y^2/(2g)
Explanation:
At maximum height, the vertical velocity becomes zero. Using v_y^2 = u_y^2 - 2gH with v_y = 0 gives H = u_y^2/(2g).
2. At the maximum height of a projectile, its vertical component of velocity is:
A.
Maximum
B.
Equal to g
C.
Zero
D.
Equal to its horizontal component
Correct Answer:
Zero
Explanation:
The projectile stops moving upward momentarily at maximum height, so its vertical velocity is zero.
3. A projectile is launched with a vertical velocity component of 20 m/s. If g = 10 m/s², what is its maximum height?
A.
10 m
B.
20 m
C.
30 m
D.
40 m
Correct Answer:
20 m
Explanation:
The maximum height is H = u_y^2/(2g) = 20^2/(2 × 10) = 20 m.
4. If the initial vertical component of a projectile's velocity is doubled, its maximum height becomes:
A.
Twice the original height
B.
Half the original height
C.
Four times the original height
D.
Unchanged
Correct Answer:
Four times the original height
Explanation:
Maximum height is proportional to the square of the initial vertical velocity, H = u_y^2/(2g). Doubling u_y makes the height four times larger.
5. A projectile has an initial vertical velocity component of 30 m/s. Taking g = 10 m/s², what is its maximum height?
A.
15 m
B.
30 m
C.
45 m
D.
60 m
Correct Answer:
45 m
Explanation:
Using H = u_y^2/(2g), H = 30^2/(2 × 10) = 45 m.
6. Which equation is most directly used to calculate the maximum height of a projectile when its initial vertical velocity is known?
A.
H = u_y^2/(2g)
B.
H = u_y/g
C.
H = 2u_y/g
D.
H = gu_y^2/2
Correct Answer:
H = u_y^2/(2g)
Explanation:
At maximum height, the final vertical velocity is zero. Applying the vertical motion equation gives H = u_y^2/(2g).
7. Two projectiles are launched with the same initial vertical velocity component on the same planet. Their maximum heights are:
A.
Different in all cases
B.
Equal
C.
Inversely proportional to their horizontal velocities
D.
Always zero
Correct Answer:
Equal
Explanation:
Maximum height depends on the initial vertical velocity component and gravitational acceleration, not on the horizontal velocity. Since both are the same, the maximum heights are equal.
8. A projectile reaches a maximum height of 5 m. If g = 10 m/s², what was its initial vertical velocity component?
A.
5 m/s
B.
10 m/s
C.
15 m/s
D.
20 m/s
Correct Answer:
10 m/s
Explanation:
From H = u_y^2/(2g), u_y = √(2gH) = √(2 × 10 × 5) = 10 m/s.
9. At maximum height, the vertical acceleration of a projectile is:
A.
Zero
B.
Equal to the horizontal velocity
C.
Directed upward with magnitude g
D.
Directed downward with magnitude g
Correct Answer:
Directed downward with magnitude g
Explanation:
Even at maximum height, gravity continues to act downward. Therefore, the vertical acceleration remains downward with magnitude g.
10. If the initial vertical velocity component of a projectile is zero, its maximum height above the point of projection is:
A.
Zero
B.
g
C.
Equal to the horizontal velocity
D.
Infinite
Correct Answer:
Zero
Explanation:
With no initial upward vertical velocity, the projectile does not rise above its point of projection, so its maximum height is zero.