The height h of an object projected upward from the floor of the Golden Gate Bridge which has an initial height of 220 feet and with an initial velocity of 30 feet per second is given by h=-16t^2+30t+220. How many seconds later will the rock hit the water? Round to the nearest tenth. Please help ASAP!!! :(
Assuming that the 200 feet is the height above the water,
-16t^2+30t+220 = 0
8t^2 - 15t - 110 = 0
Use your favourite method of solving a quadratic, reject the
negative answer.
To find the time it takes for the rock to hit the water, we need to determine when the height (h) is equal to zero. We can do this by setting the height equation equal to zero and solving for t.
The height equation is:
h = -16t^2 + 30t + 220
Setting h to zero, we have:
0 = -16t^2 + 30t + 220
At this point, we need to solve this quadratic equation to find the values of t when the rock hits the water.
There are different methods to solve quadratic equations, but one common method is using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 30, and c = 220.
Plugging these values into the quadratic formula, we get:
t = (-30 ± √(30^2 - 4*(-16)*220)) / (2*(-16))
Simplifying further, we have:
t = (-30 ± √(900 + 14080)) / (-32)
t = (-30 ± √(14980)) / (-32)
Evaluating √(14980) ≈ 122.43 (rounded to two decimal places), we have:
t ≈ (-30 ± 122.43) / (-32)
Now, considering both the positive and negative square root, we have two possible solutions:
t₁ ≈ (-30 + 122.43) / (-32)
t₂ ≈ (-30 - 122.43) / (-32)
Calculating these values, we get:
t₁ ≈ 0.84
t₂ ≈ 4.39
Since time cannot be negative in this context, we discard the negative value t₂. Therefore, the time it takes for the rock to hit the water is approximately 0.84 seconds (rounded to the nearest tenth).