Suppose an equation that describes the path of a diver when diving off a platform is d = -5t^2 + 10t + 20. Where d is the distance above the water in (feet) and t is the time from the beginning of the dive(seconds). After how many seconds is the diver 25 feet above the water ? After how many seconds does the diver enter the water?
Solve for t in d(t)=25 and d(t)=0, where d(t) is given as:
d(t)= -5t^2 + 10t + 20
we find that d(1)=25, and d(3.23)=0.
Note: not sure if the diver was on earth. The acceleration due to gravity is only 10 instead of 32.2 f/s2 unless the distance d is in metres.
To find the time when the diver is 25 feet above the water, we need to set the equation equal to 25 and solve for t:
-5t^2 + 10t + 20 = 25
To solve this quadratic equation, we can rearrange it:
-5t^2 + 10t - 5 = 0
Now we can apply the quadratic formula: t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -5, b = 10, and c = -5. Plugging these values into the formula:
t = (-10 ± √(10^2 - 4(-5)(-5))) / (2(-5))
t = (-10 ± √(100 - 100)) / (-10)
t = (-10 ± √(0)) / (-10)
t = (-10 ± 0) / (-10)
t = 0 / -10 or t = 0 / -10
Since t cannot be divided by zero, we know that at t = 0 seconds, the diver is 25 feet above the water.
Now let's find when the diver enters the water. To do that, we need to set the equation equal to zero and solve for t:
-5t^2 + 10t + 20 = 0
We can't solve this equation algebraically, so we'll need to use the quadratic formula again. Plugging in the values of a = -5, b = 10, and c = 20:
t = (-10 ± √(10^2 - 4(-5)(20))) / (2(-5))
t = (-10 ± √(100 + 400)) / (-10)
t = (-10 ± √(500)) / (-10)
Since we have a positive value inside the square root, there are two possible solutions:
t = (-10 + √(500)) / (-10) or t = (-10 - √(500)) / (-10)
Calculating the square root of 500, we get:
t = (-10 + 22.36) / (-10) or t = (-10 - 22.36) / (-10)
t = 12.36 / (-10) or t = -32.36 / (-10)
t = -1.236 or t = 3.236
Since time cannot be negative in this context, the diver enters the water at t ≈ 3.236 seconds.