A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after the launch will the rocket be 500 ft above the ground? Round to the nearest hundredth of a seconds?
-16t2 + 250t = 500
solve for t
t=2
No.
You have to set this equal to zero.
16t^2 -250t + 500 = 0
Now, this isn't factorable. You have to use the quadratic equation to solve for t to the nearest hundredth.
Quadratic x = opposite of b (plus or minus) the square root of( b squared minus 4ac). All of this divided by 2a.
To find the number of seconds after the launch when the rocket will be 500 ft above the ground, we need to solve the equation h = 500.
The equation that represents the height of the rocket at any given time is h = -16t^2 + 250t. We can substitute 500 for h in this equation:
500 = -16t^2 + 250t.
Now, we need to rearrange this equation to solve for t. Let's bring all terms to one side of the equation:
16t^2 - 250t + 500 = 0.
To solve this quadratic equation, we can use the quadratic formula, which is given by:
t = (-b ± √(b^2 - 4ac)) / 2a,
where a = 16, b = -250, and c = 500.
Plugging in these values, we get:
t = (-(-250) ± √((-250)^2 - 4 * 16 * 500)) / (2 * 16),
Simplifying further:
t = (250 ± √(62500 - 32000)) / 32,
t = (250 ± √30500) / 32.
Now, we can calculate the two possible solutions for t:
t1 = (250 + √30500) / 32,
t2 = (250 - √30500) / 32.
To round to the nearest hundredth of a second, we can use a scientific calculator or an online calculator.
Calculating t1 ≈ 7.18 seconds,
Calculating t2 ≈ 15.82 seconds.
Therefore, the rocket will be 500 ft above the ground approximately 7.18 seconds and 15.82 seconds after the launch.