A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s=-16t^2+208t. After how many seconds will the ball be 640 feet from the ground?
To find the time at which the ball will be 640 feet from the ground, we can set the equation for the distance of the ball from the ground equal to 640 and solve for t.
The equation for the distance of the ball from the ground is given as s = -16t^2 + 208t.
Substituting s = 640, we get:
640 = -16t^2 + 208t
To solve this quadratic equation, we need to set it equal to zero:
-16t^2 + 208t - 640 = 0
Now, we can solve this quadratic equation by factoring or by using the quadratic formula.
Factoring:
Start by finding two numbers that multiply to give -640 (the product of the coefficient of t^2 and the constant term) and add up to give 208 (the coefficient of t):
Let's use the numbers -32 and 20.
-32 * 20 = -640
-32 + 20 = -12
Using these numbers, we can rewrite the quadratic equation:
-16t^2 + 208t - 640 = 0
-16t^2 - 12t + 20t - 640 = 0
Now, group the terms and factor by grouping:
-4t(4t - 5) + 32(4t - 5) = 0
From this, we can see that (4t - 5) is a common factor. Factoring it out, we get:
(4t - 5)(-4t + 32) = 0
Now, we can set each factor equal to zero and solve for t:
4t - 5 = 0 --> 4t = 5 --> t = 5/4 = 1.25
-4t + 32 = 0 --> -4t = -32 --> t = -32/-4 = 8
Since we are dealing with time, t cannot be negative, so we discard t = 8.
Therefore, the ball will be 640 feet from the ground after 1.25 seconds.
And that wasn't even the maximum height !!!
solve -16t^2+208t = 640
t^2 - 13t + 40 = 0
(t-5)(t-8) = 0
t = .....
interpret this result