one of the fireworks is launched from the top of the building with an initial upward velocity of 150 ft/sec
the initial velocity tales place of a different coefficent of "t" in the equation
The equation for the vertical motion of the fireworks can be given by the formula:
h(t) = h₀ + v₀t + (1/2)gt²
Where:
h(t) is the height of the fireworks at time t
h₀ is the initial height of the fireworks (top of the building)
v₀ is the initial velocity of the fireworks (upward velocity)
g is the acceleration due to gravity (approximately 32 ft/sec²)
In this case, since the fireworks is launched from the top of the building, the initial height (h₀) would be the height of the building. However, the question does not provide the height of the building, so we cannot determine the initial height accurately.
The initial velocity (v₀) is given as 150 ft/sec.
The coefficient of "t" in the equation represents the initial velocity (v₀).
To determine the equation for the motion of the firework, we can use kinematic equations. The equation relating the vertical displacement (y), initial velocity (v0), time (t), and acceleration due to gravity (g) is:
y = v0t - 0.5gt^2
Here, we are given the initial upward velocity of the firework, v0, as 150 ft/sec. The equation becomes:
y = 150t - 0.5gt^2
However, you mentioned that the initial velocity takes place of a different coefficient of "t" in the equation. It is unclear what specifically you meant by that statement. If you can provide additional information or clarify your question, I would be happy to assist you further.