Home
Class 11
PHYSICS
A particle is fired from the ground at s...

A particle is fired from the ground at some angle `(theta ne 90^(@))` and it returns to the ground after some time . Select the correct option (s)

A

Acceleration and velocity becomes perpendicular to each other once .

B

Direction of acceleration during first half of motion is upwards and during second half of motion it is downwards .

C

Angle between the velocity and acceleration during the first half of motion is greater than `90^(@)` and during the second half of motion the same is less than `90^(@)`.

D

Acceleration remains constant .

Text Solution

Verified by Experts

The correct Answer is:
A, C, D

(a,c,d): At the highest point of journey velocity becomes horizontal and hence perpendicular to the acceleration due to gravity . Option (a) is correct .
During the motion, acceleration remains contant and is equal to acceleration due to gravity , hence option (b) is work and (d) is correct
During the first half of motion, velocity of the particle is directed above the horizontal and during the second half, velocity of the particle is directed below the horizontal, hence we can understand that option (c) is also correct .
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    MODERN PUBLICATION|Exercise COMPETITION FILE OBJECTIVE TYPE QUESTIONS (D. MULTIPLE CHOICE QUESTIONS)|9 Videos
  • MOTION IN A PLANE

    MODERN PUBLICATION|Exercise COMPETITION FILE OBJECTIVE TYPE QUESTIONS (Assertion Reason Type Questions)|10 Videos
  • MOTION IN A PLANE

    MODERN PUBLICATION|Exercise COMPETITION FILE OBJECTIVE TYPE QUESTIONS (JEE (Advanced) for IIT Entrance)|2 Videos
  • MECHANICAL PROPERTIES OF FLUIDS

    MODERN PUBLICATION|Exercise Chapter Practise Test|16 Videos
  • MOTION IN A STRAIGHT LINE

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST|16 Videos

Similar Questions

Explore conceptually related problems

A particle is fired from a point on the ground with speed u making an angle theta with the horizontal.Then

A particle is projected from the ground at an angle such that it just clears the top of a polar after t_1 time its path. It takes further t_2 time to reach the ground. What is the height of the pole ?

In the problem 150, the time taken to return to the ground from the maximum height:

A particle is thrown vertically upward from the ground with some velocity and it strikes the ground again in time 2 s. The maximum height achieved by the particle is : (g=10 m//s^(2))

A particle is projected from ground with velocity 40sqrt2m at 45^(@) .At time t=2s

A particle is projected from ground with speed 80 m/s at an angle 30^(@) with horizontal from ground. The magnitude of average velocity of particle in time interval t = 2 s to t = 6 s is [Take g = 10 m//s^(2)]

A particle is projected from surface of the inclined plane with speed u and at an angle theta with the horizontal. After some time the particle collides elastically with the smooth fixed inclined plane for the first time and subsequently moves in vertical direction. Starting from projection, find the time taken by the particle to reach maximum height. (Neglect time of collision).

A particle is executing vertical SHM about the highest point of a projectile. When the particle is at the mean position, the projectile is fired from the ground with velocity u at an angle theta with the horizontal.The projectile hits the oscillating particle .Then, the possible time period of the particle is