A volcano erupts, and a chunk of hot magma is launched horizontally with a speed of 208\,\dfrac{\text m}{\text s}208 s m 208, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction from a height of 94\,\text m94m94, start text, m, end text. We can ignore air resistance.

a. 42.92 m/s b 4.4 s c. 915.2 m

Explanation:

Here is the complete question

A volcano erupts, and a chunk of hot magma is launched horizontally with a speed of   208 m/s, from a height of 94 m. We can ignore air resistance.

What is the vertical velocity of the magma when it hits the ground?

Solution

a. Using v² = u² - 2gy since it is motion under gravity, where u = initial vertical velocity = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and y =  height = 94 m.

Substituting these values into the equation, we have

v² = u² - 2gy

v² = 0² - 2 × (-9.8 m/s²) × 94 m

v² = 1842.4 m²/s²

v = √1842.4 m²/s²

v = 42.92 m/s

b. We also find the time it takes to drop from the height of 94 m from

y = ut - 1/2gt² , u = 0

So, t = √(2y/g)

= √(2 × 94 m ÷ 9.8 m/s²)

= √19.184 s²

= 4.4 s

c. The horizontal distance the chunk of hot magma covers is thus x = Vt where v = horizontal velocity = 208 m/s

x = 208 m/s × 4.4 s = 915.2 m

the answer is c.)

explanation:

it may be wrong but i think its a