68.A body is projected horizontally from a height of 78.4m with a velocity of 10m/s Its velocity after 3seconds is (g=10m/s2)(Take direction of projection as I and vertically upward direction as j)

Open in App

Solution

$Velocity after 3 seconds : \vec{v} = \vec{u} + \vec{a} t = \vec{u} + \vec{g} t = 10 \hat{i} + 10 (- \overset{⏜}{j}) (3) = (10 \hat{i} - 30 \overset{⏜}{j}) m / s$

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 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

Q. Body is projected vertically with velocity

10

m/s on reaching the ground, change in its velocity is(upwards direction is taken as positive)

Q. A stone is projected horizontally with speed

10 m / s

from a height

h

above the ground. A horizontal wind is blowing in the direction opposite to the velocity of projection and gives the stone a constant horizontal acceleration

10 m / s^{2}

(in the direction opposite to the initial velocity). As a result the stone falls on ground at a point vertically below the point of projection. Then the value of height

h

is (

g = 10 m / s^{2}

)

Q. A body is projected with a velocity of 20 m/s in a direction making an angle of

60^{\circ}

with the horizontal. Calculate
(i) its position after 0.55
(ii) velocity after 0.55

Join BYJU'S Learning Program

68.A body is projected horizontally from a height of 78.4m with a velocity of 10m/s Its velocity after 3seconds is (g=10m/s2)(Take direction of projection as I and vertically upward direction as j)

Velocity after 3 seconds:v→=u→+a→t =u→+g→t =10i^+10-j⏜3 =10i^-30j⏜ m/s

$Velocity after 3 seconds : \vec{v} = \vec{u} + \vec{a} t = \vec{u} + \vec{g} t = 10 \hat{i} + 10 (- \overset{⏜}{j}) (3) = (10 \hat{i} - 30 \overset{⏜}{j}) m / s$