Home
Class 11
PHYSICS
A bullet is to be fired with a speed of ...

A bullet is to be fired with a speed of `2000 m//s` to hit a target `200 m` away on a level ground. If `g = 10 m//s^(2)`, the gun should be aimed

A

Directly at the target

B

5 cm below the target

C

5 cm above the target

D

2 cm above the target

Text Solution

Verified by Experts

The correct Answer is:
C
Time taken by the bullet to hit target
`t = 200/2000 = 1/10` sec.
Vertical height
`y = h = 1/2 "gt"^(2) = 1/2 xx 10 xx (1/10)^(2) =1/20 m = 5 cm [u_(v) =0]`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN TWO DIMENSION

    ERRORLESS|Exercise NCERT BASED QUESTIONS (OBLIQUE PROJECTILE MOTION)|47 Videos

  • MOTION IN TWO DIMENSION

    ERRORLESS|Exercise PAST YEARS QUESTIONS |22 Videos

  • MOTION IN TWO DIMENSION

    ERRORLESS|Exercise NCERT BASED QUESTIONS (NON-UNIFORM CIRCULAR MOTION)|36 Videos

  • MOTION IN ONE DIMENSION

    ERRORLESS|Exercise ASSERTION AND REASON|24 Videos

  • NEWTON'S LAWS OF MOTION

    ERRORLESS|Exercise ASSERTION & REASON |18 Videos

Similar Questions

Explore conceptually related problems

A bullet is fired with a speed of 1000 m/sec in order to penetrate a target situated at 100 m away. If g = 10 m//s^(2) , the gun should be aimed

A bullet is fired with a speed of 10^(3)ms^(-1) in order to hit s target 100m away. If g= 10ms^(-2) , the gun should be aimed at

A bullet is fired from a gun eith a speed of 1000m/s on order to hit target 100 m away At what height above target should the gun be aimed ? (The resistance of air is negligible and g = 10 m//s^(2) )

A bullet of mass 50 kg is fired with a speed of 200 m//s from a gun of mass 2 kg . Find the recoil velocity of the gun.

A bullet is fired from a cannon with velocity 500 m/s. If the angle of projection is 15^(@) and g = 10m//s^(2) . Then the range is

A ball with projection velocity 10 m//s is to hit a target 6 m away in the same horizontal line. How high above the target must the ball be aimed so that the ball will hit the target ? (tan 18.5^(@) = 1//3, tan 71.5^(@) = 3)