Home
Class 12
PHYSICS
A projectile is thrown with a velocity o...

A projectile is thrown with a velocity of `10sqrt(2) ms^(-1)` at an angle of 45° with the horizontal. The time interval between the moments when the speeds are `sqrt(125) ms^(-1)` is (g=10`ms^(-2)`)

Text Solution

Verified by Experts


`u_(n) = 10, u_(y)=0`
`v^(2) =v_(x)^(2) + v_(y)^(2)`
`125 = 100 +v_(y)^(2)`
`Deltat =(2v_(y))/g =(2 xx 5)/10 = 1s`
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise SHORT ANSWER TYPE QUESTIONS|6 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise VERY SHORT ANSWER TYPE QUESTIONS|10 Videos

  • MOTION IN A PLANE

    AAKASH SERIES|Exercise EXERCISE-3 (Circular Motion)|7 Videos

  • MOTION IN A STRAIGHT LINE

    AAKASH SERIES|Exercise very Short answer type question|15 Videos

Similar Questions

Explore conceptually related problems

A body is projected from height of 60 m with a velocity 10 ms^(-1) at angle 30° to horizontal. The time of flight of the body is [g = 10ms^(-2) ]

A body is projected with velocity 24 ms^(-1) making an angle 30° with the horizontal. The vertical component of its velocity after 2s is (g=10 ms^(-1) )

A body is projected with velocity 24 ms^(-1) making an angle 30° with the horizontal. The angle made by the direction of the projectile with the horizontal at 2s from start is

A bullet is projected upwards from the top of a tower of height 90 m with the velocity 30 ms^(-1) making an angle 30° with the horizontal. The time taken by it to reach the ground is (g=10 ms^(-2) )

A body is projected from height of 60 m with a velocity 10 ms^(-1) at angle 30^(@) to horizontal The time of flight of the body is [g =10ms^(-2)]

A ball is thrown with velocity 8 ms^(-1) making an angle 60° with the horizontal. Its velocity will be perpendicular to the direction of initial velocity of projection after a time of (g =10ms^(-2) )