Question

A bomb at rest at the Summit of a cliff breaks into two equal fragments one of the fragments attains a horizontal velocity 20√3 m/s the horizontal distance between the two fragments when the displacement vectors incline 60° related to each other is

Answer 1

Explanation:

20

20 3

20 3

20 3 )t=d

20 3 )t=dS=0+

20 3 )t=dS=0+ 2

20 3 )t=dS=0+ 21

20 3 )t=dS=0+ 21

20 3 )t=dS=0+ 21 ×gt

20 3 )t=dS=0+ 21 ×gt 2

20 3 )t=dS=0+ 21 ×gt 2

20 3 )t=dS=0+ 21 ×gt 2 S=5t

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 =

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = s

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd =

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 =

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12s

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3 ×12

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3 ×12So, 2d=480

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3 ×12So, 2d=480 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3 ×12So, 2d=480 3

20 3 )t=dS=0+ 21 ×gt 2 S=5t 2 Now, tan30 0 = sd = 5t 2 20 3 t 3 1 = t4 3 t=12sd=20 3 ×12So, 2d=480 3 m