A body is projected horizontally with a velocity of 4 metre per second from the top of a high tower. The velocity of the body after 0.7 seconds is nearly
(take $g = 10 m / s^{2}$ )

10m/s

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

8m/s

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

19.2m/s

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

11m/s

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

Open in App

Solution

The correct option is D 8m/s
Horizontal componenet ux of velocity is unaffected.
At the starting point, the vertical component of velocity is zero.
After 0.7 s vertical component of velocity uy is given by, uy = g×t = 10×0.7 = 7 m/s
velocity of the body after 0.7s = $\sqrt{4^{2} + 7^{2}} = 8 m / s$

Suggest Corrections

Similar questions

Q. A body is projected horizontally from the top of a tower with

30 m s^{- 1}

. If

g = 10 m s^{- 2}

, the velocity of that body after

4

seconds is

Q. A body is projected horizontally with a velocity of

4 \sqrt{2}

m/sec. The velocity of the body after 0.7 seconds will be nearly: (Take

g = 10 m / s e c^{2}

)

Q. A body is projected horizontally from the top of a high tower with a speed of

20 m s^{- 1}

. After

4

seconds, the displacement of the body is

(g = 10 m s^{- 2})

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

Join BYJU'S Learning Program

A body is projected horizontally with a velocity of 4 metre per second from the top of a high tower. The velocity of the body after 0.7 seconds is nearly(take g=10m/s2)

The correct option is D 8m/sHorizontal componenet ux of velocity is unaffected.At the starting point, the vertical component of velocity is zero.After 0.7 s vertical component of velocity uy is given by, uy = g×t = 10×0.7 = 7 m/svelocity of the body after 0.7s = √42+72=8m/s

A body is projected horizontally with a velocity of 4 metre per second from the top of a high tower. The velocity of the body after 0.7 seconds is nearly
(take $g = 10 m / s^{2}$ )