A ball is thrown straight up from a bridge over a river and falls into the water. The height, h, in meters, of the ball above the water t seconds after being thrown is approximately modeled by the relation h=-5t^2+10t+35.
c) After how many seconds does the ball hit the water?
since h is the height above the water, it hits the water when h=0:
-5t^2+10t+35 = 0
t = 1+2√2
Where did you get the 1 + 2 square root 2 for the answer? Please explain more thoroughly for the answer and how you got it.
To find out when the ball hits the water, we need to determine the time at which the height (h) of the ball is equal to zero. In other words, we need to solve the equation -5t^2 + 10t + 35 = 0.
To solve this quadratic equation, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = -5, b = 10, and c = 35. Plugging these values into the quadratic formula, we have:
t = (-10 ± √(10^2 - 4*(-5)*35)) / (2*(-5))
Simplifying further:
t = (-10 ± √(100 + 700)) / (-10)
t = (-10 ± √800) / (-10)
Now, we need to simplify the square root. 800 can be written as 16 * 50, and the square root of 16 is 4. Therefore, √800 = 4√50.
t = (-10 ± 4√50) / (-10)
Now, we can split the solutions into two separate cases:
Case 1: t = (-10 + 4√50) / (-10)
Case 2: t = (-10 - 4√50) / (-10)
Calculating these values will give us two possible answers for the time when the ball hits the water.