Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 202227 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
27 July
Shift
Morning Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2022 · 27 July · Morning Shift · Physics · Motion in One Dimension

A bullet is shot vertically downwards with an initial velocity of...

A bullet is shot vertically downwards with an initial velocity of 100ms−1100m{s}^{- 1} from a certain height. Within 10s10s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity-time curve for total time t=20st = 20s will be : (Take g=10ms−2g = 10m{s}^{- 2})A bullet is shot vertically downwards with an initial velocity of 100m s ^ - 1 from a certain height. Within 10s , theA bullet is shot vertically downwards with an initial velocity of 100m s ^ - 1 from a certain height. Within 10s , theA bullet is shot vertically downwards with an initial velocity of 100m s ^ - 1 from a certain height. Within 10s , theA bullet is shot vertically downwards with an initial velocity of 100m s ^ - 1 from a certain height. Within 10s , the

Options

  1. ALet the maximum and minimum value of the expression 2 ^ 2 θ + cos θ + 1 is M and m respectively, then the value of
  2. BLet the maximum and minimum value of the expression 2 ^ 2 θ + cos θ + 1 is M and m respectively, then the value of
  3. CLet the maximum and minimum value of the expression 2 ^ 2 θ + cos θ + 1 is M and m respectively, then the value of
  4. DLet the maximum and minimum value of the expression 2 ^ 2 θ + cos θ + 1 is M and m respectively, then the value of

Correct answer

A. Let the maximum and minimum value of the expression 2 ^ 2 θ + cos θ + 1 is M and m respectively, then the value of

Step-by-step solution

As the upward direction is taken as positive, the velocity at t=0t = 0 is u=−100ms−1u = - 100m{s}^{- 1}.

As it is falling under gravity, its acceleration is constant and a=−10ms−2a = - 10m{s}^{- 2}.

Applying equation of motion for constant acceleration during the time of fall i.e., for 00 to 10s10s,

v=u+at⇒v=−100−10×10⇒v=−200ms−1v = u + at ⇒ v = - 100 - 10 \times 10 ⇒ v = - 200m{s}^{- 1}

After 10s10s it hits the ground and remains at rest.

Therefore, option A is the correct representation.

Practice Motion in One Dimension on Quantrex Academy →

More from Motion in One Dimension

A gas balloon is going up with a constant velocity of 10 m/s. When this balloon reached a… JEE Main 2026Two masses of 3.4 kg and 2.5 kg are accelerated from an initial speed of 5 m/s and 12… JEE Main 2026The velocity ( v ) versus time ( t ) plot of a particle is shown in the figure, for a… JEE Main 2026From 18 m height above the ground a ball is dropped from rest. The height above the… JEE Main 2026Two cars A and B are moving in the same direction along a straight line with speeds 100… JEE Main 2026The velocity of a particle is given as v = -x i + 2y j - z k m/s. The magnitude of… JEE Main 2026A particle starts moving from time t=0 and its coordinate is given as x(t)=4 t^ 3 -3 t A.… JEE Main 2026Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the… JEE Main 2026 Full Motion in One Dimension list All JEE Main PYQs