The answer to this question is <span>13,537</span>
Answer:
The value is 
Explanation:
From the question we are told that
The horizontal speed is 
The horizontal distance is 
Generally the time taken by the hot magma in air before landing is mathematically represented as

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

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

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

=> 
B is the correct answer for sure bro
Answer:
i don't know if this is good for you but
Explanation:
ignoring frictional air resistance (drag) the speed on return is the same as when it left the ground (5 m/s but in the opposite direction).
Note: this points out a good reason for not firing live bullets into the air..they will return somewhere and at the same speed.
However, if you take into account the atmospheric drag the reurn speed will be somewhat smaller (but in the case of a bullet, probably still lethal.) Drag depends on many factors and is difficult to calculate.
FREQUENCY is the number of complete waves that pass a given point in a certain amount of time.
Good luck :)