Question 56060: 1. A person dives from a 30 foot platform with an initial velocity of
> 5 feet per second (upward)
>
> a. Find the position of the function
> b. How long will it take for the diver to hit the water?
> c. What is the diver's velocity at impact?
> d. What is the diver's acceleration at impact?
Answer by jenrobrody(19)   (Show Source):
You can put this solution on YOUR website! For a falling body problem, use the function:
where a represents the acceleration due to gravity= -32 ft/sec^2
b represents initial velocity= 5
c represents initial height= 30
(notice that up is positive and down is negative here)
the function f(x)= -16x^2+ 5x + 30, gives the position of the diver at time t.
Note that if x=0, f(0) = 30. ie, the diver is 30 feet above water at t=0.
We need to find time=x when f(x)=0. ie, at zero height above water.
Sovle for x:
x = -49.1/-32 or 39.1/-32
x = 1.5 or -1.2 seconds
Use the positive answer for time after jumping.
For the velocity, use the equation(same constants a,b,c as above):
or g(x)=(-32)x+5 = -32x + 5
where g(x) represents the velocity after x seconds.
Find velocity after 1.5 seconds:
g(1.5) = -32(1.5) + 5 = -43 feet/sec
For acceleration, h(x)= a or, h(x)= -32 feet/sec^2
acceleration is a constant and equals -32 feet/sec^2.
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