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A rocket is launched upward with an initial velocity of 1960 m/s. After how many minutes does it fall to the ground? (in meter the formula h=rt-4.9t^2 is used

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5 votes

Answer:

3.33 minutes

Explanation:

h = rt - 4.9t^2

In this case, the rocket is launched upward, so the initial velocity (r) is positive 1960 m/s. The rocket will fall to the ground when the height (h) becomes zero.

0 = 1960t - 4.9t^2

To solve this quadratic equation, we can set it equal to zero and use the quadratic formula:

4.9t^2 - 1960t = 0

Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 4.9, b = -1960, and c = 0. Plugging in these values:

t = (-(-1960) ± √((-1960)^2 - 4 * 4.9 * 0)) / (2 * 4.9)

Simplifying further:

t = (1960 ± √(3841600)) / 9.8

t = (1960 ± 1960) / 9.8

t = 1960 / 9.8 = 200

So, the rocket will fall to the ground after 200 seconds. To convert this to minutes, divide by 60:

200 seconds / 60 = 3.33 minutes (approximately)

Therefore, the rocket will fall to the ground after approximately 3.33 minutes.

answered
User Alex Gao
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