A model rocket is launched with an initial velocity of 200 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 200t.
How many seconds after the launch will the rocket be 550 ft above the ground? Round to the nearest hundredth of a second. (Enter your answers as a comma-separated list.)
To find the time when the rocket will be 550 ft above the ground, we need to solve the equation:
h = -16t^2 + 200t
We set h to 550 and solve for t:
550 = -16t^2 + 200t
Moving all terms to one side, we get:
16t^2 - 200t + 550 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -200, and c = 550. Plugging these values into the formula, we have:
t = (200 ± √((-200)^2 - 4 * 16 * 550)) / (2 * 16)
Calculating the discriminant, we get:
√((-200)^2 - 4 * 16 * 550) = √(40000 - 35200) = √4800 = 69.28
Now we can plug this value into the formula:
t = (200 ± 69.28) / 32
Using the plus-minus sign, we have two values for t:
t₁ = (200 + 69.28) / 32 = 269.28 / 32 = 8.42 seconds
t₂ = (200 - 69.28) / 32 = 130.72 / 32 = 4.08 seconds
Rounding to the nearest hundredth of a second, the rocket will be 550 ft above the ground approximately 4.08 seconds and 8.42 seconds after the launch. Therefore, the answer is:
4.08, 8.42
To find the number of seconds after the launch when the rocket will be 550 ft above the ground, we need to solve the equation:
h = -16t^2 + 200t
Given that the height is 550 ft, we can substitute h with 550 and solve for t.
550 = -16t^2 + 200t
To solve this quadratic equation, we can rearrange it to standard form:
-16t^2 + 200t - 550 = 0
Now, we can use the quadratic formula to find the values of t:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 200, and c = -550.
t = (-200 ± sqrt(200^2 - 4(-16)(-550))) / (2(-16))
Now, we can simplify and calculate the roots using a calculator or a math software:
t = (-200 ± sqrt(40000 - 35200)) / (-32)
t = (-200 ± sqrt(4800)) / (-32)
t = (-200 ± 69.28) / (-32)
Now, we can calculate the two possible values for t:
t1 = (-200 + 69.28) / (-32) ≈ 4.47 seconds
t2 = (-200 - 69.28) / (-32) ≈ 14.53 seconds
Therefore, the rocket will be approximately 550 ft above the ground at around 4.47 seconds and 14.53 seconds after the launch.
You want -16t^2 + 200t to be equal to 550, so ...
-16t^2 + 200t = 550
16t^2 - 200t + 550 = 0
Use the quadratic formula, you will get 2 positive answers.
Of course the smaller time would be on it way up, the larger time
will be the time on its return.