Since math and physics are the same I was wondering If I could get help with this question.
Q1).An athlete executes a long jump and leaves the ground at an angle of 33 degrees and travels 7.8m . What is the take-off speed?
Q2). A projectile is fired with an initial speed of 51.2m/s at an angle 44.5 degrees above the horizonon a flat firering range, Determine.
a). Maximum hieght reached by the projectile
b). total time in the air.
c). the range covered
d). the velocity after 1.5sec.
Q3). Determine how much further a person can jump on the moon then on earth , if the take-off speed and angle are the same. The acceleration due to gravity on the moon is one-sixth what it is on earth.
This should get you going. All of the following is undoubtedly in your physics book.
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.
