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3 years ago

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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?

1 answer:

ArbitrLikvidat [17]3 years ago

7 0

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

$- v_y = 48.61 \ m/s$

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

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