Kinematics-2D
127 Questions
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Q126
Objective
4. Match the columns
Numerical
A particle is projected horizontally from a tower with velocity $10 \, ms^{-1}$ . Taking, $g = 10 \, ms^{-2}$ . Match the following two columns at time t = 1s and mark the correct option from the codes given below.
| Column I | Column II | |||||||
| (A) | Horizontal component of velocity | (p) | 5 SI units | |||||
| (B) | Vertical component of velocity | (q) | 10 SI unit | |||||
| (C) | Horizontal displacement | (r) | 15 SI unit | |||||
| (D) | Vertical displacement | (s) | 20 SI unit | |||||
| Codes | ||||||||
| A | B | C | D | A | B | C | D | |
| (a) | p | q | s | r | (b) | q | s | p |
| (c) | q | p | r | s | (d) | q | q | p |
Correct Answer: D
Explanation:
In horizontal projectile motion,
Horizontal component of velocity,
$u _ {x} = u = 1 0 \mathrm{ms} ^ {- 1}$
Vertical component of velocity,
$u _ {y} = g t = 1 0 \times 1 = 1 0 \mathrm{ms} ^ {- 1}$
Horizontal displacement
$= u \times t = 1 0 \times (1) = 1 0 \mathrm{m}$
Vertical displacement = $\frac{1}{2}gt^{2}=\frac{1}{2}\times10\times(1)^{2}=5m$
Horizontal component of velocity,
$u _ {x} = u = 1 0 \mathrm{ms} ^ {- 1}$
Vertical component of velocity,
$u _ {y} = g t = 1 0 \times 1 = 1 0 \mathrm{ms} ^ {- 1}$
Horizontal displacement
$= u \times t = 1 0 \times (1) = 1 0 \mathrm{m}$
Vertical displacement = $\frac{1}{2}gt^{2}=\frac{1}{2}\times10\times(1)^{2}=5m$
Q127
Objective
5. Collection of questions asked in NEET & various medical entrance exams
MCQ
A block is dragged on a smooth plane with the help of a rope which moves with a velocity v as shown in the figure. The horizontal velocity of the block is
A.
$\frac{v}{\sin\theta}$
B.
$v \sin \theta$
C.
$\frac{v}{\cos\theta}$
D.
$v\cos \theta$
