Practice important
Trajectory Equation
MCQ questions and answers.
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MCQ Questions
1. What is the equation of the trajectory of a projectile projected with speed u at an angle θ to the horizontal?
A.
y = x tanθ - gx²/(2u² cos²θ)
B.
y = x tanθ + gx²/(2u² cos²θ)
C.
y = x cotθ - gx²/(2u² sin²θ)
D.
y = x tanθ - gx²/(2u² sin²θ)
Correct Answer:
y = x tanθ - gx²/(2u² cos²θ)
Explanation:
The trajectory equation for a projectile is y = x tanθ - gx²/(2u² cos²θ), obtained by eliminating time from its horizontal and vertical motion equations.
2. In the trajectory equation y = x tanθ - gx²/(2u² cos²θ), what does x represent?
A.
Vertical displacement of the projectile
B.
Horizontal displacement of the projectile
C.
Initial speed of the projectile
D.
Acceleration due to gravity
Correct Answer:
Horizontal displacement of the projectile
Explanation:
In the trajectory equation, x denotes the horizontal distance traveled by the projectile from its point of projection.
3. In the equation y = x tanθ - gx²/(2u² cos²θ), what does y represent?
A.
Horizontal displacement
B.
Initial vertical velocity
C.
Vertical displacement
D.
Horizontal velocity
Correct Answer:
Vertical displacement
Explanation:
The symbol y represents the vertical displacement or height of the projectile above the point of projection.
4. The trajectory equation y = x tanθ - gx²/(2u² cos²θ) shows that the path of a projectile is a:
A.
Straight line
B.
Circle
C.
Parabola
D.
Hyperbola
Correct Answer:
Parabola
Explanation:
Since the trajectory equation contains an x² term, it is a quadratic equation in x and therefore represents a parabola.
5. Which term in the trajectory equation y = x tanθ - gx²/(2u² cos²θ) causes the path to curve downward?
A.
x tanθ
B.
tanθ
C.
gx²/(2u² cos²θ)
D.
u² cos²θ
Correct Answer:
gx²/(2u² cos²θ)
Explanation:
The negative quadratic term, -gx²/(2u² cos²θ), produces the downward curvature of the projectile's parabolic path.
6. If the projection angle θ is zero, what does the trajectory equation become?
A.
y = x
B.
y = -gx²/(2u²)
C.
y = gx²/(2u²)
D.
y = u²x/g
Correct Answer:
y = -gx²/(2u²)
Explanation:
For θ = 0, tanθ = 0 and cosθ = 1. Substitution gives y = -gx²/(2u²), the trajectory of a horizontally projected particle.
7. At the point of projection, where x = 0, what value does the trajectory equation give for y?
A.
y = u
B.
y = g
C.
y = 0
D.
y = tanθ
Correct Answer:
y = 0
Explanation:
Putting x = 0 in y = x tanθ - gx²/(2u² cos²θ) makes both terms zero, so y = 0 at the point of projection.
8. Which quantity appears in the denominator of the quadratic term in the projectile trajectory equation?
A.
2u² cos²θ
B.
2u cosθ
C.
u² sinθ
D.
g cos²θ
Correct Answer:
2u² cos²θ
Explanation:
The standard trajectory equation contains the quadratic term gx²/(2u² cos²θ), so its denominator is 2u² cos²θ.
9. For a projectile launched with a fixed speed and angle, if gravity is neglected, what form does its trajectory equation take?
A.
y = x tanθ
B.
y = -x tanθ
C.
y = gx²/(2u² cos²θ)
D.
y = x² tanθ
Correct Answer:
y = x tanθ
Explanation:
When g = 0, the quadratic term vanishes, leaving y = x tanθ, which represents a straight-line path.
10. In the trajectory equation y = x tanθ - gx²/(2u² cos²θ), which factor determines the initial slope of the path?
A.
g
B.
tanθ
C.
u²
D.
cos²θ
Correct Answer:
tanθ
Explanation:
The coefficient of the first-degree term x is tanθ, so it determines the initial slope of the trajectory at the point of projection.