1. A particle is projected at an angle θ\thetaθ to the horizontal with
initial velocity uuu. Which of the following expressions represents the
time of flight in terms of uuu, ggg, and θ\thetaθ?
A. 2ugsinθ\frac{2u}{g} \sin\thetag2usinθ
B. 2ugcosθ\frac{2u}{g} \cos\thetag2ucosθ
C. 2usinθg\frac{2u\sin\theta}{g}g2usinθ
D. ugsinθ\frac{u}{g} \sin\thetagusinθ
Answer: C) 2usinθg\frac{2u\sin\theta}{g}g2usinθ
Rationale: The time of flight for projectile motion is given by the
formula T=2usinθgT = \frac{2u \sin\theta}{g}T=g2usinθ, where
uuu is the initial speed, θ\thetaθ is the angle of projection, and ggg is
the acceleration due to gravity.
2. A ball is thrown horizontally with a speed of 12 m/s from a height of
5 m. How far will it travel horizontally before hitting the ground?
(Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 20 m
B. 24 m
C. 30 m
D. 35 m
Answer: B) 24 m
,Rationale: First, find the time to fall: t=2hg=2×510=1=1 st =
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1
\, \text{s}t=g2h=102×5=1=1s.
Then, the horizontal distance is d=u×t=12×1=12 md = u \times t = 12
\times 1 = 12 \, \text{m}d=u×t=12×1=12m.
3. A block slides on a rough surface with a coefficient of friction
μ=0.2\mu = 0.2μ=0.2. If the block has a mass of 5 kg, what is the force
of friction acting on it?
A. 10 N
B. 20 N
C. 25 N
D. 50 N
Answer: A) 10 N
Rationale: The frictional force is given by
Ffriction=μmgF_{\text{friction}} = \mu mgFfriction=μmg, where
μ=0.2\mu = 0.2μ=0.2, m=5 kgm = 5 \, \text{kg}m=5kg, and
g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2. So,
Ffriction=0.2×5×10=10 NF_{\text{friction}} = 0.2 \times 5 \times
10 = 10 \, \text{N}Ffriction=0.2×5×10=10N.
4. A particle is moving along a straight line with constant velocity.
What is its acceleration?
A. 0 m/s²
B. 1 m/s²
, C. 9.8 m/s²
D. Undefined
Answer: A) 0 m/s²
Rationale: If the particle is moving with constant velocity, there is no
change in speed or direction, meaning the acceleration is zero.
5. A ball is thrown vertically upwards with an initial velocity of 30 m/s.
What will be the time taken to reach the maximum height? (Take
g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 1.5 s
B. 2.5 s
C. 3.0 s
D. 5.0 s
Answer: B) 3.0 s
Rationale: The time to reach maximum height is given by t=ugt =
\frac{u}{g}t=gu, where u=30 m/su = 30 \, \text{m/s}u=30m/s and
g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2. So, t=3010=3 st =
\frac{30}{10} = 3 \, \text{s}t=1030=3s.
6. A body is moving in a circular path of radius rrr. The angular
velocity of the body is ω\omegaω. What is the relationship between
the linear velocity vvv and angular velocity ω\omegaω?
A. v=ωrv = \omega rv=ωr
B. v=rω2v = r \omega^2v=rω2
initial velocity uuu. Which of the following expressions represents the
time of flight in terms of uuu, ggg, and θ\thetaθ?
A. 2ugsinθ\frac{2u}{g} \sin\thetag2usinθ
B. 2ugcosθ\frac{2u}{g} \cos\thetag2ucosθ
C. 2usinθg\frac{2u\sin\theta}{g}g2usinθ
D. ugsinθ\frac{u}{g} \sin\thetagusinθ
Answer: C) 2usinθg\frac{2u\sin\theta}{g}g2usinθ
Rationale: The time of flight for projectile motion is given by the
formula T=2usinθgT = \frac{2u \sin\theta}{g}T=g2usinθ, where
uuu is the initial speed, θ\thetaθ is the angle of projection, and ggg is
the acceleration due to gravity.
2. A ball is thrown horizontally with a speed of 12 m/s from a height of
5 m. How far will it travel horizontally before hitting the ground?
(Take g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 20 m
B. 24 m
C. 30 m
D. 35 m
Answer: B) 24 m
,Rationale: First, find the time to fall: t=2hg=2×510=1=1 st =
\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1
\, \text{s}t=g2h=102×5=1=1s.
Then, the horizontal distance is d=u×t=12×1=12 md = u \times t = 12
\times 1 = 12 \, \text{m}d=u×t=12×1=12m.
3. A block slides on a rough surface with a coefficient of friction
μ=0.2\mu = 0.2μ=0.2. If the block has a mass of 5 kg, what is the force
of friction acting on it?
A. 10 N
B. 20 N
C. 25 N
D. 50 N
Answer: A) 10 N
Rationale: The frictional force is given by
Ffriction=μmgF_{\text{friction}} = \mu mgFfriction=μmg, where
μ=0.2\mu = 0.2μ=0.2, m=5 kgm = 5 \, \text{kg}m=5kg, and
g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2. So,
Ffriction=0.2×5×10=10 NF_{\text{friction}} = 0.2 \times 5 \times
10 = 10 \, \text{N}Ffriction=0.2×5×10=10N.
4. A particle is moving along a straight line with constant velocity.
What is its acceleration?
A. 0 m/s²
B. 1 m/s²
, C. 9.8 m/s²
D. Undefined
Answer: A) 0 m/s²
Rationale: If the particle is moving with constant velocity, there is no
change in speed or direction, meaning the acceleration is zero.
5. A ball is thrown vertically upwards with an initial velocity of 30 m/s.
What will be the time taken to reach the maximum height? (Take
g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2)
A. 1.5 s
B. 2.5 s
C. 3.0 s
D. 5.0 s
Answer: B) 3.0 s
Rationale: The time to reach maximum height is given by t=ugt =
\frac{u}{g}t=gu, where u=30 m/su = 30 \, \text{m/s}u=30m/s and
g=10 m/s2g = 10 \, \text{m/s}^2g=10m/s2. So, t=3010=3 st =
\frac{30}{10} = 3 \, \text{s}t=1030=3s.
6. A body is moving in a circular path of radius rrr. The angular
velocity of the body is ω\omegaω. What is the relationship between
the linear velocity vvv and angular velocity ω\omegaω?
A. v=ωrv = \omega rv=ωr
B. v=rω2v = r \omega^2v=rω2
