Practice important
Trajectory Equation
MCQ questions and answers.
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MCQ Questions
1. Which equation represents 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 = ut + 1/2 gt²
Correct Answer:
y = x tanθ - gx²/(2u² cos²θ)
Explanation:
The trajectory equation of 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 vertical velocity
D.
Time of flight
Correct Answer:
Horizontal displacement of the projectile
Explanation:
In the trajectory equation, x denotes the horizontal distance travelled by the projectile from its point of projection.
3. In the trajectory equation y = x tanθ - gx²/(2u² cos²θ), what does y represent?
A.
Horizontal range
B.
Initial speed
C.
Vertical height of the projectile
D.
Horizontal component of velocity
Correct Answer:
Vertical height of the projectile
Explanation:
The symbol y represents the vertical displacement or height of the projectile above the point of projection.
4. The trajectory equation of a projectile is generally an equation of which type?
A.
A straight line
B.
A circle
C.
A parabola
D.
An ellipse
Correct Answer:
A parabola
Explanation:
The trajectory equation contains a term proportional to x², so the path of a projectile is a parabola.
5. For a projectile projected horizontally, what is the form of its trajectory equation?
A.
y = x
B.
y = gx²/(2u²)
C.
y = x tanθ
D.
y = u²x²/(2g)
Correct Answer:
y = gx²/(2u²)
Explanation:
For horizontal projection, θ = 0, so tanθ = 0 and cosθ = 1. Substitution into the trajectory equation gives y = -gx²/(2u²) when upward y is positive; using downward vertical displacement gives y = gx²/(2u²).
6. In the equation y = x tanθ - gx²/(2u² cos²θ), which term causes the projectile path to curve?
A.
x tanθ
B.
y
C.
gx²/(2u² cos²θ)
D.
u² cos²θ
Correct Answer:
gx²/(2u² cos²θ)
Explanation:
The term containing x² makes the trajectory equation quadratic in x, producing the curved parabolic path.
7. If the angle of projection is 45°, the trajectory equation becomes which expression?
A.
y = x - gx²/u²
B.
y = x + gx²/u²
C.
y = x - gx²/(2u²)
D.
y = x/2 - gx²/u²
Correct Answer:
y = x - gx²/u²
Explanation:
For θ = 45°, tan45° = 1 and cos²45° = 1/2. Therefore, y = x - gx²/[2u²(1/2)] = x - gx²/u².
8. At the point of projection, what are the values of x and y in the trajectory equation?
A.
x = 1 and y = 1
B.
x = 0 and y = 0
C.
x = u and y = g
D.
x = g and y = u
Correct Answer:
x = 0 and y = 0
Explanation:
The origin is chosen at the point of projection, so both horizontal and vertical displacements are zero there.
9. Which variable is eliminated to obtain the trajectory equation from the horizontal and vertical motion equations?
A.
Horizontal distance
B.
Time
C.
Acceleration due to gravity
D.
Angle of projection
Correct Answer:
Time
Explanation:
The trajectory equation is obtained by eliminating time between x = u cosθ t and y = u sinθ t - 1/2 gt².
10. For a fixed initial speed and projection angle, how does the height y depend on the horizontal distance x in the trajectory equation?
A.
It depends linearly only on x
B.
It depends quadratically on x
C.
It is independent of x
D.
It depends only on time
Correct Answer:
It depends quadratically on x
Explanation:
The trajectory equation includes the term -gx²/(2u² cos²θ), showing that y has a quadratic dependence on x.