Uhhhh I don't think that the correct answer.
First, find the gradient of the line segment.
Gradient = Rise/Run = (2-6)/(2-5) = 4/3.
Since the other line segment is parallel to this line segment, their gradients are the same.
That means (-2+4)/(x-5) = 4/3.
2/(x-5) = 4/3
It is obvious x = 6.5
Answer:
31.9secs
6,183.3m
Step-by-step explanation:
Given the equation that models the height expressed as;
h(t ) = -4.9t²+313t+269
At the the max g=height, the velocity is zero
dh/dt = 0
dh/dt = -9,8t+313
0 = -9.8t + 313
9.8t = 313
t = 313/9.8
t = 31.94secs
Hence it takes the rocket 31.9secs to reach the max height
Get the max height
Recall that h(t ) = -4.9t²+313t+269
h(31.9) = -4.9(31.9)²+313(31.9)+269
h(31.9) = -4,070.44+9,984.7+269
h(31.9) = 6,183.3m
Hence the maximum height reached is 6,183.3m
(4 sec) * (-10 ft/sec) = -40 ft.
I believe your answer is 53
