A volcano launches a lava bomb straight upward with an initial speed of 28 m/s. Taking upward to the positive direction, find the speed and direction of motion of the lava bomb

(a)v = 8.4 m/s(upward)

(b)v=1.4 m/s (downward)

Explanation:

Given,

Initial velocity of volcano,u = 28 m/s

We have to find the speed and direction of the volcano bomb at 2 sec and 3 sec.

Now,

Using the equation of motion

v = u + g t

g = -9.8 m/s²

At t = 2 s

v = 28 - 9.8 x 2

v = 8.4 m/s

So, velocity of lava at t=2 sec is 8.4 m/s in upward direction.

At t= 3 sec

v = 28 - 9.8 x 3

v = -1.4 m/s

So, velocity of lava at t= 3 sec is 1.4 m/s in downward direction.

b

explanation:

bc its just b

overcharge is the right answer system