Question

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 direction of first body w.r.t second is?

Answer 1

In case one, u = 10 m/s, v = 0, acceleration = A1 and let time = 1 sec (constant)

Therefore, a = v – u/t = 0 – 10/1 m/sec sq. = -10 m/sec sq.

In case two, u = 0, v = 10m/s Similarly, A2 = 10 - 0/1 = 10 m/sec sq.

Now, magnitude of second body with respect to first body = 10-10 = zero.

Answer 2

Answer:

zero

Explanation:

u=0 , v=0, a= A1, t=1 m/s^2, g= 9.8 m/s^2

for upward-   u= 0, v=10, then

A1= v-u/t m/s^2

 = 10-0/1 m/s^2

 = 10 m/s^2  

for downward- u= 10, v= o, a= A2,  then

A2=v-u/t m/s^2

   =0-10/1 m/s^2

  =  -10 m/s^2

magnitude of second body with respect to first body = A1+A2

                                                                 = 10+(-10)m/s^2

                                                                = 10-10 m/s^2

                                                               = 0