Home
Class 11
PHYSICS
Assertion Particle-1 is dropped from a t...

Assertion Particle-1 is dropped from a tower and particle-2 is projected horizontal from the same tower. Then both the particles reach the ground simultaneously.
Reason Both are particles strike the ground with different speeds.

A

If both Asseration and Reason are correct and Reason is the correct explanation of Assertion

B

If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion

C

If Assertion is true but Reason is false

D

If Assertion is false but Reason is true

Text Solution

Verified by Experts

The correct Answer is:
B

`t_(1) = t_(2) = sqrt((2h)/(g))` (h = height of tower)
`v_(1) = sqrt(2gh)`
While `v_(2) = sqrt(v_(1)^(2)+v_(0)^(2)) (v_(0)=` initial horizontal velocity)
Therefore, both assertion and reason are correct but reason is not the correct explanation of assertion.
Promotional Banner

Topper's Solved these Questions

  • MOTION

    DC PANDEY|Exercise Match the Columns|4 Videos
  • MOTION

    DC PANDEY|Exercise C. Medical entrances gallery|1 Videos
  • MOTION

    DC PANDEY|Exercise B. Medical entrance|1 Videos
  • MEASUREMENT AND ERRORS

    DC PANDEY|Exercise Subjective|19 Videos
  • MOTION IN A PLANE

    DC PANDEY|Exercise (C )Medical entrances gallery|32 Videos

Similar Questions

Explore conceptually related problems

A ball is dropped from the top of an 80 m high tower After 2 s another ball is thrown downwards from the tower. Both the balls reach the ground simultaneously. The initial speed of the second ball is

Height of a tower is 125m and a particle is dropped from rest from the top of the tower. After two seconds of its fall, another particle is projected downwards with a speed u such that both the particles reach the ground simultaneously. What is the value of u?

A particle is projected horizontally from a tower with velocity u towards east. Wind provides constant horizontal acceleration (g)/(2) towards west. If particle strikes the ground vertically the then :

A particle is dropped from the top of a tower. It covers 40 m in last 2s. Find the height of the tower.

A particle is projected from a tower as shown in figure, then find the distance from the foot of the tower where it will strike the ground. (g=10m//s^(2))

Two particles A and B are placed as shown in figure. The particle A, on the top of tower, is projected horizontally with a velocity u and particle B is projected along the surface towards the tower, simultaneously. If both particles meet each other, then the speed of projection of particles B is [ignore any friction]

A particle is dropped from a height h. Another particle which is initially at a horizontal distance d from the first is simultaneously projected with a horizontal velocity u and the two particles just collide on the ground. Then

A particle is dropped from a height h .Another particle which is initially at a horizontal distance d from the first is simultaneously projected with a horizontal velocity u and the two particles just collide on the ground.Then

A particle is projected at angle theta with horizontal from ground. The slop (m) of the trajectory of the particle varies with time (t) as

A particle is thrown horizontally from a tower of height 20m . If the speed of projection is 20m/s then the distance of point from the tower,where the particle hit the ground is:-