You visited us 1 times! Enjoying our articles? Unlock Full Access!

Potential Energy of an Object at a Height

A body is pro...

Question

A body is projected horizontally from a height of $78.4 m$ with a velocity $10 {m s}^{- 1}$ . Its velocity after $3$ seconds is ____ ( $g = 10 {m s}^{- 1}$ ). (Take direction of projection on $^i$ and vertically upward direction on $^j$ ):

10^i - 30^j

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10^i + 30^j

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20^i - 30^j

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10^i + 10 \sqrt{30}^j

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $10^i - 30^j$
Initial horizontal velocity $u_{x} = 10 m / s$
Initial vertical velocity $u_{y} = 0 m / s$
Vertical velocity at $t = 3$ seconds
$v_{y} = u_{y} - g t$
$= 0 - 10 \times 3 = - 30 m / s$
$= - 30^j m / s$
Velocity after 3 seconds: $\to v = 10^i - 30^j$

Suggest Corrections

Similar questions

A body is projected horizontally from a height of 78.4m with a velocity 10 M/s. Its velocity after 3seconds is ? (G=10m/S2)(take the direction of projection on i vector and vertically upward direction on J vector)

Q. A body is projected vertically upwards with a velocity of

10 {m s}^{- 1}

and another body is projected simultaneously from the same point with a velocity of

20 {m s}^{- 1}

at an angle of

\frac{π}{6}

with the horizontal. The distance between the two bodies after one second from the time of projection is (Acceleration due to gravity is

10 {m s}^{- 2}

)

Q. A body projected vertically upward with a velocity

v

t = 0

is found at a height

h

after

1

second and after further

6

seconds, it is found at the same height (

g = 10 m s^{- 2}

). Then,

Q. A force

\to F = 4^i - 5^j + 3^k

is acting at point

_{1} =^i + 2^j + 3^k

. The torque acting about the point

_{2} = 3^i - 2^j - 3^k

Q. At a height of

15 m

from ground velocity of a projectile

\to v = (10^i + 10^j)

and (

g = 10 m s^{- 2}

)

Join BYJU'S Learning Program

A body is projected horizontally from a height of 78.4m with a velocity 10ms−1. Its velocity after 3 seconds is ____ (g=10ms−1). (Take direction of projection on ^i and vertically upward direction on ^j):

The correct option is A 10^i−30^jInitial horizontal velocity ux=10m/sInitial vertical velocity uy=0m/sVertical velocity at t=3secondsvy=uy−gt=0−10×3=−30m/s=−30^jm/sVelocity after 3 seconds:→v=10^i−30^j

A body is projected horizontally from a height of $78.4 m$ with a velocity $10 {m s}^{- 1}$ . Its velocity after $3$ seconds is ____ ( $g = 10 {m s}^{- 1}$ ). (Take direction of projection on $^i$ and vertically upward direction on $^j$ ):

The correct option is A $10^i - 30^j$
Initial horizontal velocity $u_{x} = 10 m / s$
Initial vertical velocity $u_{y} = 0 m / s$
Vertical velocity at $t = 3$ seconds
$v_{y} = u_{y} - g t$
$= 0 - 10 \times 3 = - 30 m / s$
$= - 30^j m / s$
Velocity after 3 seconds: $\to v = 10^i - 30^j$