Bench Mark Kinematics use the four Kinematic
Equations to describe things being launched into the
air at an angle. The Kinematic equations are V=Vo+at, x=1/2(V+Vo)t,
x=Vot+1/2at^2, V^2=Vo^2+2ax, where V=Final Velocity,
Vo= Initial Velocity, x=distance, t=time, and a=accelerations.
Example
A foot ball is punted in the air for 3.8 seconds. Solve for the
height and initial velocity.
First, you must decide which equation to use. You are given time,
but when solving for the height, you must divide time by 2 because you are only trying to
find the information at its apex, and the ball reaches at half the time. So time would be
1.9 seconds while solving for the height. When solving for the initial velocity, just use
the original time. Acceleration, while the ball is going up, is -9.8 m/s^2 because of gravity.
Final velocity is 0 because when the ball reaches its highest point in the air,
its velocity 0 for just a split second just when it stops going up and starts going
down, and you only want to find how fast the ball is going at its highest point
So, you have time, final velocity, and acceleration. You can use V=Vo+at
to solve for the initail velocity, then use x=Vot+1/2at^2. So, the work would
look like this:
V=Vo+at
0=Vo+(-9.8)3.8
0=Vo+(-37.24)
Vo=37.24
Now that you have initial velocity, you can solve for the height using x=1/2(V+Vo)t.
Remember to half the time because you only want to find the ball half way through its course.
x=1/2(V+Vo)t
x=1/2(0+37.24)1.9
x=1/2(37.24)1.9
x=35.378
So the initial velocity equals 37.24 m/s and the height equals 35.378 m.
Let's try one more.
A 54-kg shell is launched from a 21 degree angle into the air.
Its initial velocity is 88 m/s. Assuming it lands at the same elevation it was launched from,
how far did it travel, and how high did it go?
First, you need to find the initial velocity. However, since you are
given the angle, there will be 2 different velocities: one for the shell going across, and
one for the shell going up. To find these, use the equations x=V(cos of the angle) for
distance and y=(sin of the angle) for height.
x=88(cos 21)x=82.15
y=88(sin 21)y=31.54
Now that you have the initial velocities, you can solve for the time. Use the initial
velocity for the shell going across so you get the full time, not half the time.
V=Vo+at
0=82.16+(-9.8)t
-82.16=(-9.8)t
t=8.38
Now, with the time, you can solve for the distance and height. We will solve for the
distance first.
x=1/2(V+Vo)t
x=1/2(0+82.16)8.38
x=344.25 m
Now to solve for the height. Remember to half the time. However, you can just use the
basic distance equation: x=vt
x=vt
x=31.54(4.19)
x=132.15
So distance=344.25 meters and height=132.15 meters.
And that is how you can use the kinematic equations to solve bench
mark equations.
