The height h(t) measured in feet of an object dropped by an astronaut on the moon can be approximated by h(t) = h0 (0 on the bottom) - 2.7t^2, where h0 is the height from which the object was dropped. About how long would it take an object to fall to the moon (h = 0) if it were dropped by an astronaut from a height of 6 feet?
so
0 = -2.7t^2 + 6
t^2 = 6/2.7
t = √(6/2.7) or appr 1.49 seconds
To find the time it takes for the object to fall to the moon (h = 0), we need to set h(t) = 0 in the given equation:
0 = h0 - 2.7t^2
We are given h0 = 6 feet, so the equation becomes:
0 = 6 - 2.7t^2
Now we need to solve this equation for t. Let's rearrange the equation:
2.7t^2 = 6
Divide both sides by 2.7:
t^2 = 6 / 2.7
t^2 = 2.222...
To find t, take the square root of both sides:
t = √2.222...
Calculating the square root, we get:
t ≈ 1.49
Therefore, it would take approximately 1.49 seconds for the object to fall to the moon if it were dropped by an astronaut from a height of 6 feet.
To find out how long it would take for an object to fall to the moon (h = 0) when dropped from a height of 6 feet, we can set the equation h(t) = 0 and solve for t.
The equation given is h(t) = h0 - 2.7t^2, where h(t) represents the height of the object at time t and h0 is the initial height from which the object was dropped.
Substituting the given values, we have:
0 = 6 - 2.7t^2
Now, let's solve this quadratic equation to determine the value of t.
Rearranging the equation, we get:
2.7t^2 = 6
Dividing both sides by 2.7, we have:
t^2 = 6 / 2.7
Simplifying further, we get:
t^2 = 2.2222
To find the value of t, we take the square root of both sides:
√(t^2) = √2.2222
This gives us:
t ≈ √2.2222
Using a calculator, we find:
t ≈ 1.489 seconds (approximately)
Therefore, it would take approximately 1.489 seconds for an object dropped by an astronaut from a height of 6 feet to fall to the moon (h = 0).