trig
posted by Anonymous .
Bob throws a ball straight up with an initial speed of 50 feet per second from a height of 6 feet.
(a) Find parametric equations that describe the motion of the ball as a function of time. (Do this on paper. Your instructor may ask you to turn in this work.)
(b) How long is the ball in the air?
seconds
(c) When is the ball at its maximum height?
seconds
Determine the maximum height of the ball.
feet
(d) Simulate the motion of the ball by graphing the equations found in part (a).

trig 
tchrwill
Read this over and you should be able to solve your problem.
The acceleration of a body is defined as the change of its velocity during an interval of time divided by the interval of time. Expressed algebraically, a = (Vo  Vf)/t where a = the acceleration, Vo = the initial velocity, Vf = the final velocity and t = the interval of time.
Uniformly Accelerated Motion
Three equations present the relationships between the initial velocity, the final velocity, the distance covered, acceleration and time.
The first expresses the relationship between the initial and final velocities and a time interval.
From the basic a = (Vo  Vf) we derive the first as
......................................t
..............................Vf = Vo + at (the acceleration is assumed constant)
The second defines the relationship between the distance traveled by a body exposed to a constant acceleration. The distance traveled over an interval of time derives from the product of the average velocity during the time interval and the time interval. The average velocity is (Vo + Vf)/2 and the time interval "t" which yields d = (Vo + Vf)t/2. Substituting the Vf derived above, d = [Vo + (Vo + at)]t/2 resulting in
...............................d = Vo(t) + a(t^2) (quite often, s is used in place of d)
......................................................2
The third combines the first two by eliminating the time interval "t".
Therefore, a(d) = (Vo  Vf)(Vo + Vf)t = (1/2)(Vf  Vo)(Vf + Vo) resulting in
.....................................t.............2
..............................Vf^2 = Vo^2 + 2ad
As written, these three equations apply to the uniformly accelerated motion of a body in a horizontal plane.
Another application of these equations is to falling bodies under the influence of the earth's gravity. All bodies fall to the ground due to the force of gravity pulling on them. As strange as it may seem, all bodies fall with the same acceleration, regardless of their masses. The acceleration of a falling body, caused by the force of gravity, depends on the distance of the body from the center of the earth. The gravitational force, and thereby the acceleration, decreases the further the body is from the center of the earth. The acceleration due to gravity is given the notaion "g" and averages 32.17 ft./sec.^2. Quite often, for solving instructional problems, g = 32 ft./sec.^2 is acceptable.
The equations of accelerated motion derived earlier also apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g). This results in
..............................Vf = Vo + gt (the acceleration is assumed constant)
...............................d = Vo(t) + g(t^2)
.....................................................2
..............................Vf^2 = Vo^2 + 2gd
As written, these expressions apply to falling bodies. The equations that apply to rising bodies are
..............................Vf = Vo  gt (the acceleration is assumed constant)
...............................d = Vo(t)  g(t^2)
.....................................................2
..............................Vf^2 = Vo^2  2gd
All of the above ignores surface and atmospheric friction.
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