Question

A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance of 1250 m before it hits the ground. We can ignore air resistance. What is the vertical velocity of the magma when it hits the ground?

Answer 1

The value is
`v_y = -48.61 \ m/s`

Step-by-step explanation:

From the question we are told that

The horizontal speed is
`u_x = 252 \ m/s`

The horizontal distance is
`d = 1250 \ m`

Generally the time taken by the hot magma in air before landing is mathematically represented as

`t = (d)/(u_x)`

=>
`t = ( 1250 )/(252)`

=>
`t = 4.96 \ s`

Generally the initial vertical velocity of the magma when it was lunched is

`u_y = 0 \ m/ s`

Then the final velocity of the magma when it hits the ground is mathematically represented s

`- v_y = u_y + gt`

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So

`- v_y = 48.61 \ m/s`

=>
`v_y = -48.61 \ m/s`