A diver jumps from a diving board that is 32 feet above the water. The height of the diver is given by
h= -16(t-2)(t+1)
where the height h is measured in feet, and the time t is measured in seconds. When will the diver hit the water?
solve
h = -16(t-2)(t+1) = 0
There two solution, t = -1 and t = +2 s.
In real situation with correct physics, the equation would be
h = 32 - 16t^2 = 16(2 - t^2)
= 16(t + sqrt2)(sqrt2 - t) and solutions would bedifferent.
To find when the diver hits the water, we need to set the height h equal to 0 since the height of the water is at the surface level. So, we have:
-16(t-2)(t+1) = 0
Since a product of factors is equal to zero if and only if one or more of the factors is zero, we can set each factor equal to zero and solve for t.
Setting t-2 = 0, we get t = 2.
Setting t+1 = 0, we get t = -1.
Since time cannot be negative in this context, we discard t = -1 as a solution.
Therefore, the diver will hit the water at t = 2 seconds.
To find when the diver will hit the water, we need to determine the value of t when the height (h) of the diver is equal to 0.
Given the equation:
h = -16(t-2)(t+1)
Setting h to 0:
0 = -16(t-2)(t+1)
Since the product of two factors is equal to zero only if one or both of the factors are equal to zero, we can set each factor equal to zero and solve for t.
For t-2 = 0:
t = 2
For t+1 = 0:
t = -1
So, the diver will hit the water at two different times: t = 2 seconds and t = -1 seconds.
However, time cannot be negative in this context, so we discard t = -1 seconds. Thus, the diver will hit the water at t = 2 seconds.