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A firecracker is shot straight up into the air with an initial velocity of 20 m/s. Find its height (y) after 2 sec (a = -10 m/s²)

User Mmilleruva
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1 Answer

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Final answer:

The height of the firecracker after 2 seconds is 0 meters.

Step-by-step explanation:

When a firecracker is shot straight up into the air with an initial velocity of 20 m/s, we can use the equation v² = v₀² - 2a(y - y₀) to find its height after 2 seconds. In this equation, v is the final velocity, v₀ is the initial velocity, a is the acceleration, y is the final height, and y₀ is the initial height.

Given that the acceleration due to gravity (a) is -10 m/s² and the initial velocity (v₀) is 20 m/s, we can substitute these values into the equation, along with the time (t) of 2 seconds, to solve for the height (y).

Using the formula, we get:

y = y₀ + v₀t + (1/2)at²

Plugging in the values, we have:

y = 0 + (20 m/s)(2 s) + (1/2)(-10 m/s²)(2 s)²

Simplifying the equation gives:

y = 20 m + (-20 m)

So, the height of the firecracker after 2 seconds is 0 meters.

User Dan Parker
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