A body is projected vertically upward with speed 10 m / s and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t second is . take .g=10 m / s 2

Question

A body is projected vertically upward with speed 10 m / s and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t second is . take .g=10 m / s 2  (A) Zero (B) 10 m / s 2 (C) 5 m / s 2 (D) 20 m / s 2

Tardigrade Admin · Accepted Answer

The acceleration of first body a1=10 ms -2 a2=10 ms -2 a rel =a1- a 2=10 ms -2-10 ms -2=0

Q. A body is projected vertically upward with speed $10 m / s$ and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t second is ${$ take $g = 10 m / s^{2}}$

Zero

$10 m / s^{2}$

$5 m / s^{2}$

$20 m / s^{2}$

The acceleration of first body
$a_{1} = 10 m s^{- 2}$
$a_{2} = 10 m s^{- 2}$
$a_{re l} = a_{1} - a_{2} = 10 m s^{- 2} - 10 m s^{- 2} = 0$

Q. A body is projected vertically upward with speed 10m/s and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t second is { take g=10m/s2}

Zero

10m/s2

5m/s2

20m/s2

The acceleration of first body a1​=10ms−2 a2​=10ms−2 arel​=a1​−a2​=10ms−2−10ms−2=0

Q. A body is projected vertically upward with speed $10 m / s$ and other at same time with same speed in downward direction from the top of a tower. The magnitude of acceleration of first body w.r.t second is ${$ take $g = 10 m / s^{2}}$

$10 m / s^{2}$

$5 m / s^{2}$

$20 m / s^{2}$

The acceleration of first body
$a_{1} = 10 m s^{- 2}$
$a_{2} = 10 m s^{- 2}$
$a_{re l} = a_{1} - a_{2} = 10 m s^{- 2} - 10 m s^{- 2} = 0$