If a rock falls from a height of 50
meters on Earth, the height H (in meters) after x seconds is approximately H(x)=50-4.9x^2
1. What is the height of the rock when x=1.6 seconds?
2. When is the height of the rock 5 meters?
3. When does the rock strike the ground?
I think the first one is 3.03 not sure if I'm right.
1. h=50-4.9*1.6^2
no, you are wrong.
thank you how do I set the the third question to get the answer.
h is zero when it hits the ground
h=50-4.9t^2
t= sqrt (50-h) / 4.9
1. 30.4
To solve these questions, we need to substitute the given values of x into the equation H(x) = 50 - 4.9x^2. Let's go through each question step by step:
1. What is the height of the rock when x = 1.6 seconds?
To find the height when x = 1.6, we substitute x = 1.6 into the equation H(x) = 50 - 4.9x^2:
H(1.6) = 50 - 4.9(1.6)^2
H(1.6) = 50 - 4.9(2.56)
H(1.6) = 50 - 12.544
H(1.6) = 37.456
So when x = 1.6 seconds, the height of the rock is approximately 37.456 meters.
2. When is the height of the rock 5 meters?
To find when the height is 5 meters, we set H(x) = 5 and solve for x:
5 = 50 - 4.9x^2
4.9x^2 = 50 - 5
4.9x^2 = 45
x^2 = 45/4.9
x^2 ≈ 9.1837
Taking the square root of both sides:
x ≈ ±3.03
So the height of the rock is 5 meters approximately at x = 3.03 seconds.
3. When does the rock strike the ground?
The rock strikes the ground when its height, H(x), is equal to 0. So we set H(x) = 0 and solve for x:
0 = 50 - 4.9x^2
4.9x^2 = 50
x^2 = 50/4.9
x^2 ≈ 10.2041
Taking the square root of both sides:
x ≈ ±3.19
So the rock strikes the ground approximately at x = 3.19 seconds.