1. What is the vertical velocity of a projectile at its maximum height?
A.
Maximum upward velocity
B.
Zero
C.
Equal to its horizontal velocity
D.
Equal to acceleration due to gravity
Correct Answer:
Zero
Explanation:
At maximum height, the upward vertical component of velocity becomes zero before the projectile starts descending.
2. The maximum height of a projectile depends on which component of its initial velocity?
A.
Only the horizontal component
B.
Only the vertical component
C.
Both components equally
D.
Neither component
Correct Answer:
Only the vertical component
Explanation:
Maximum height is determined by the initial vertical component of velocity, since gravity acts vertically.
3. The formula for the maximum height H of a projectile launched with initial speed u at angle theta is:
A.
H = u cos theta / g
B.
H = u squared sin squared theta / 2g
C.
H = u squared cos squared theta / 2g
D.
H = 2u squared sin theta / g
Correct Answer:
H = u squared sin squared theta / 2g
Explanation:
The maximum height is given by H = u² sin²(theta) / 2g, where u sin(theta) is the initial vertical velocity.
4. If the initial vertical component of a projectile's velocity is doubled, its maximum height becomes:
A.
Half as much
B.
Twice as much
C.
Four times as much
D.
Unchanged
Correct Answer:
Four times as much
Explanation:
Maximum height is proportional to the square of the initial vertical velocity, so doubling it makes the height four times larger.
5. A projectile is launched with speed u at an angle of 90 degrees to the horizontal. Its maximum height is:
A.
u squared / 2g
B.
u / g
C.
u squared / g
D.
2u squared / g
Correct Answer:
u squared / 2g
Explanation:
At 90 degrees, the entire initial speed is vertical, so the maximum height is H = u² / 2g.
6. At maximum height, the acceleration of a projectile is:
A.
Zero
B.
g upward
C.
g downward
D.
Equal to the horizontal velocity
Correct Answer:
g downward
Explanation:
Gravity continues to act downward even when the projectile reaches its maximum height.
7. A projectile is launched with an initial vertical velocity of 20 m/s. Taking 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:
Using H = v_y² / 2g, H = 20² / (2 × 10) = 20 m.
8. For a fixed initial speed, the maximum height of a projectile is greatest when it is projected at:
A.
0 degrees
B.
30 degrees
C.
45 degrees
D.
90 degrees
Correct Answer:
90 degrees
Explanation:
Maximum height depends on sin²(theta), which has its greatest value at 90 degrees.
9. If the angle of projection is increased from 30 degrees to 60 degrees while the initial speed remains the same, the maximum height:
A.
Decreases to zero
B.
Remains the same
C.
Becomes twice as large
D.
Becomes four times as large
Correct Answer:
Remains the same
Explanation:
The maximum height depends on sin²(theta). Since sin²(30 degrees) = sin²(60 degrees) = 1/4, the heights are equal.
10. A projectile has an initial vertical velocity of 10 m/s. Taking g = 10 m/s², what is its maximum height above the point of projection?
A.
2 m
B.
5 m
C.
10 m
D.
20 m
Correct Answer:
5 m
Explanation:
Using H = v_y² / 2g, H = 10² / (2 × 10) = 5 m.
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