The height of a rocket in meters t seconds after it is launched is approximated by the polynomial 0.5 at2 + vt + y where a is always -9.8, v is the initial velocity, and h is the initial height. The 4400i was launched from the ground at the same time the Q99 was launched from 175 meters above the ground. After how many seconds were the rockets at the same height?
To find the time when the rockets were at the same height, we need to set their height equations equal to each other and solve for t.
The height equation for the 4400i rocket is:
h1(t) = 0.5(-9.8)t^2 + vt + 0
The height equation for the Q99 rocket is:
h2(t) = 0.5(-9.8)t^2 + vt + 175
Setting them equal to each other:
0.5(-9.8)t^2 + vt + 0 = 0.5(-9.8)t^2 + vt + 175
Now we can simplify the equation by canceling out similar terms:
0 = 175
This equation implies that 0 = 175, which is not true. Therefore, there is no time when the two rockets are at the same height.
To find out when the rockets are at the same height, we need to set their height equations equal to each other and solve for t.
The height equation for the rocket 4400i is:
h(t) = 0.5(-9.8)t^2 + v(t) + h
The height equation for the rocket Q99 is:
h(t) = 0.5(-9.8)t^2 + v(t) + h
Let's substitute the given values:
For the rocket 4400i:
h(t) = 0.5(-9.8)t^2 + v(t) + 0
For the rocket Q99:
h(t) = 0.5(-9.8)t^2 + v(t) + 175
Now we can set these two equations equal to each other:
0.5(-9.8)t^2 + v(t) + 0 = 0.5(-9.8)t^2 + v(t) + 175
Since the equations are equal, the factors in front of the t^2 and t terms must also be equal.
0.5(-9.8) = 0.5(-9.8)
v = v
Now we can drop these common terms:
0 = 175
The equation 0 = 175 is not true, which means there is no solution for when the rockets are at the same height.
Therefore, the rockets will never be at the same height during their flight.