Write an equation that represents the path of a firework that was launched from the ground with the initial velocity that you choose, choose whatever initial velocity and consider where it fits into the equation. This time you’re launching the firework from the ground to consider how that changes the equation if you’re axis of symmetry is 4.688 and your coordinates of the vertex are 4.688, 118.687 and you were launching the firework next to a 50 foot building
To find an equation representing the path of a firework launched from the ground, we can use a quadratic equation in the form of y = ax^2 + bx + c.
Since the firework is launched next to a 50-foot building, the initial height (c) would be 50.
Considering the axis of symmetry as 4.688, this value represents the x-coordinate of the vertex. Since the equation of the axis of symmetry is x = h, where h is the x-coordinate of the vertex, the value of h in this case would be 4.688.
The coordinates of the vertex are (4.688, 118.687). The x-coordinate of the vertex is equal to -b/2a, where b is the coefficient of x and a is the coefficient of x^2. Thus, -b/2a = 4.688.
To find the initial velocity, let's consider that at the highest point of the trajectory, the vertical velocity (Vy) will be 0 since the object momentarily stops before descending. We can use the kinematic equation Vy = Vy0 + gt, where g is the acceleration due to gravity and t is the time it takes to reach the highest point. The initial vertical velocity (Vy0) can be determined as Vy0 = gt, as it would take the firework the same amount of time to reach the highest point as it would take for it to fall to the ground.
Assuming the acceleration due to gravity is -32.2 ft/s^2, we can determine the value of Vy0:
0 = Vy0 + (-32.2)t
t = Vy0/32.2
The total time of flight would be twice the time it takes to reach the highest point, so the time of flight (T) would be T = 2(Vy0/32.2).
Now, we can find the equation using the provided information:
1. Vertex form equation: y = a(x - h)^2 + k, where (h, k) denotes the coordinates of the vertex
y = a(x - 4.688)^2 + 118.687
2. To find the initial velocity (v0) and acceleration (a), we need to use the kinematic equation for vertical motion:
y = v0*t + (1/2)at^2
Since the firework is launched from the ground, y=0 when t=0:
0 = v0*0 + (1/2)a*0^2
Therefore, a = -32.2 ft/s^2 (acceleration due to gravity)
3. We can find v0 using the formula v0 = at:
v0 = -32.2 * ( Vy0/32.2 )
v0 = Vy0
After determining v0, we can substitute it into the equation:
y = a(x - 4.688)^2 + 118.687
0 = v0*t - (1/2)(32.2)t^2 + 50
Therefore, the equation representing the path of the firework would be:
y = v0*t - (1/2)(32.2)t^2 + 50.