Suppose a firework that was shot from the ground next to a 50 foot building with a velocity equation of Y equals V0XT -1/2×32.2 T squared +50 and in access symmetry of 4.688 seconds and the coordinates of the vertex of (4.688, 118.687) explodes three seconds after just launched at what height old is firework, be, when it explodes consider where the value of three fit into the equation to find the height of the firework at that time show work

Question

Suppose a firework that was shot from the ground next to a 50 foot building with a velocity equation of Y equals V0XT -1/2×32.2 T squared +50 and in access symmetry of 4.688 seconds and the coordinates of the vertex of (4.688, 118.687) explodes three seconds after just launched at what height old is firework, be, when it explodes consider where the value of three fit into the equation to find the height of the firework at that time show work

Answer 1

To find the height of the firework when it explodes, we need to substitute the value of t=3 into the equation and solve for y.

Given equation: y = v0xt - (1/2)gt^2 + 50

Substituting t=3:
y = v0(3) - (1/2)(32.2)(3)^2 + 50

Simplifying:
y = 3v0 - (1/2)(32.2)(9) + 50
y = 3v0 - 145.35 + 50
y = 3v0 - 95.35

We need to find the value of v0 to calculate the height. Since we know the vertex of the parabola is (4.688, 118.687), we can substitute those values into the equation:

118.687 = 3v0(4.688) - 145.35
118.687 + 145.35 = 3v0(4.688)
264.037 = 14.064v0
v0 = 264.037 / 14.064
v0 ≈ 18.8 ft/s (rounded to one decimal place)

Now we can substitute the value of v0 back into the equation:

y = 3(18.8) - 95.35
y = 56.4 - 95.35
y ≈ -38.9 ft (rounded to one decimal place)

Therefore, the firework explodes at a height of approximately -38.9 feet. Note that the negative sign indicates that the firework has already fallen below ground level when it explodes.