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.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft