Avolcano erupts and launches a chunk of hot magma horizontally with a speed of 252\,\dfrac{\text m}{\text s}252 s m 252, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. the magma travels a horizontal distance of 1250\,\text m1250m1250, start text, m, end text before it hits the ground. we can ignore air resistance.

The final vertical velocity is -48.66 m/s,

Explanation:

The full question is:

"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? "

In order to find  the final vertical velocity, we can find first the time it takes to reach the ground,  thus if we ignore air resistance horizontally we have uniform motion.

Time it takes to hits the ground.

For uniform motion we have the following equation

We can assume the initial position where the volcano is as , so at x = 1250 we have:

Thus we can solve for t by dividing both sides by 252 m/s to get

Final vertical velocity.

With the time we can work with the vertical velocity using free fall motion

Since the chunk of magma was thrown horizontally the initial vertical velocity is 0, , so we get

Thus we get

Thus the final vertical velocity is -48.66 m/s, the negative sign tell us that it is aiming down.

d.7.5cm is your answer