An apple is thrown into the air with an initial velocity of 60 feet/sec from a height of 4 feet.
a) Write a function rule that gives the height of the apple in feet as a function of time in seconds
b) When will the apple hit the ground?
c) At what time will the apple reach its maximum height?
d) What is the maximum height the apple with reach?
Digesting the following should allow you to tackle problems of this sort:
Equations of Uniformly Accelerated Motion
Acceleration - The acceleration of a body is defined as the change in its velocity during an interval of time divided by the duration of the time interval. If Vo is the initial velocity at the beginning of the period of time and Vf is the final velocity at the end of the period of time, the change in velocity is Vo - Vf. If the velocity change occurs over the period of time t, the acceleration of the body is given by a = (Vo - Vf)/t.
The relationships between initial velocity, final velocity, distance covered and time, in uniformlay accelerated motion are defined mathematically by the following three equations:
1--From a = (Vo - Vf)/t, the final velocity of a body under constant acceleration is given by Vf = Vo + at.
2--The second equation regarding accelerated motion defines the distance traveled by a body under uniform acceleration.The average velocity of a moving body during a time interval "t" is expressed by Vav = (Vo + Vf)/2. The distance traveled, "s", during this time interval "t" is the product of the average velocity and the duraion of the time interval or s = (Vo + Vf)t/2. Substituting Vf = Vo + at into this expression yields s = [Vo + (Vo + at)]t/2 or s = Vot + at^2/2.
3--The third equation of uniformly accelerated motion is derived from the first two by eliminating the time interval "t". Multiplying the two expressions results in as = (Vf - Vo)/t x (Vo + Vf)t/2 or Vf^2 = Vo^2 + 2as.
In summary,
Vf = Vo + at
s = Vot + at^2/2
Vf^2 = Vo^2 + 2as
These same equations apply to rising and falling bodies with the exception that a is replaced by g, the acceleration due to gravity.
For rising bodys,
Vf = Vo - gt
s = Vot - gt^2/2
Vf^2 = Vo^2 - 2gs.
For falling bodys,
Vf = Vo + gt
s = Vot + gt^2/2
Vf^2 = Vo^2 + 2gs
Projectile Motion
1--The study of projectile motion is made easy by breaking the initial velocity into its vertical and horizontal components. Thus, If a projectile is fired with an initial velocity of Vo at an angle "µ" to the horizontal, Vv = Vvertical = (Vo)sin(µ) and Vh = Vhorizontal = (Vo)cos(µ). The time of flight may be obtained by summing the rise time with the fall time. From Vf = Vo - gt and Vf = 0, Vv = gt making the rise time t1 = Vv/t. During this period of time, the projectile travels horizontally d = Vht1. During the rise time, t1, the projectile rises to a height of h = Vvt1 - g(t1^2)/2 which can now be written as h = g(t1^2) - g(t1^2)/2 or h = g(t1^2)/2. Clearly, the time, t2, required for the projectile to fall back to the ground derives from -h = -g(t2^2)/2 making t1 = t2 or the total time T = 2t1.
Combining these expressions, d = Vocos(µ)t = 2Vot1cos(µ) = 2Vo[Vosin(µ)/g]cos(µ) or
d = Vo^2(sin(2µ))/2g.
Eliminating t1 from Vosin(µ) = gt1 and h = gt1^2/2 yields the maximum height reached from
Vo^2(sin^2µ) = g^2(t1)^2 and h = gt1^2/2, t1^2 = Vo^2sin^µ/g^2 = 2h/g or
h = Vo^2(sin^2(µ))/2g.
2--Envision an airplane on a bombing run flying at a constant altitude "s" and with constant horizontal velocity "Vh". The objective target "O" is being observed by the bombardier. When should the bombardier release the bomb in order to hit the target? More specifically, at what angle to the vertical should the bombardier release the bomb?
Note that when the bomb is released, it retains the same forward velocity of the airplane. Therefore, the bomb travels horizontally with a constant velocity independently of the simulaneous accelerated motion toward the earth's surface. As a result, the bomb will strike a point on the ground well ahead of the airplane's position at the time of release. The horizontal distance ahead of the airplane, "x", will make an angle "?" with the vertical. Having "s", the altitude of the airplane, "x", the horizontal distance from the local vertical to the target objective"O", and "t", the time from release to the target, we have s = gt^2/2, x = Vh(t) and x = s(tan?), combining these expressions to eliminate both "x" and "t", we derive tan? = Vhsqrt(2/gs). Therefore, the bombardier releases the bomb when his instruements indicate that the angle from his local vertical position to the target is
? = arctan[Vh(sqrt(2/gs))].
(Note that the atmospheric friction effects are neglected in the above derivations.)
