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

Answer:

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

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

       $t = \frac{d}{u_x}$

=>    $t = \frac{ 1250 }{252}$

=>    $t = 4.96 \ s$

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  

        $- v_y = 48.61 \ m/s$

=>     $v_y = -48.61 \ m/s$