PROJECTILE MOTION : Q & A FOR UNDERSTANDING

1.      A roadrunner runs directly off a cliff with an initial velocity of 3.5 m/s

a)      What are the components of this velocity?

V_x = 3.5 m/s                            V_y = 0 m/s

b)      What will be the horizontal velocity 2 seconds after the bird leaves the cliff?

3.5 m/s – horizontal velocity is unchanging

c)      If the cliff is 300 m high, at what time will the roadrunner reach the ground?

h = d_y = ½ x 10 x t² = 300

300 x 2 / 10 = t² = 60

t = 7.75 s

d)     How far from the cliff will this bird land?

d_x = 3.5 x 7.75 = 27.125 m

e)      If there is a small pond which begins 25m away from the cliff and extends 2.5 meters from there; will the roadrunner land in the pond?

Yes, the pond is from 25 m to 27.5 m, so the roadrunner will land in the pond.

f)       What is the final vertical velocity at which the roadrunner is traveling? [The vertical velocity at the time when the bird reaches the ground]

V_y = 10 x 7.75 + 0 = 77.5 m/s

g)      What is the final horizontal velocity at which the roadrunner is traveling? [The horizontal velocity at the time when the bird reaches the ground]

V_x = 0 + 3.5 = 3.5 m/s

h)      What is the total final velocity of this motion? [magnitude and direction]

V² = 77.5² + 3.5² = 6018.5

V = 77.579 m/s

q = tan^-1 (77.5 / 3.5) = 87.41^o below the horizontal

2.      An object (any object) is dropped from a height of 300m

a)      How long does it take this object to fall to the ground?

h = d_y = ½ x 10 x t² = 300

300 x 2 / 10 = t² = 60

t = 7.75 s

b)      Compare this answer with you answer from question 1, part c). What are the reasons for any similarities or differences?

They are the same because their vertical motions are identical. All objects fall with the same gravitational acceleration, so two objects at the same height with the same initial vertical velocity will reach the ground at the same time.

3.      The intent of a bean bag toss game is to get your bean bag to land on the ‘bull’s-eye’ of a target. The target is set up parallel to the ground and is the same height above the ground as your hand is when you let go of the bean bag. The game’s rules further require you to be 5 m from the center of the target when you release the bag.

a)      Evaluate the following questions for both an angle of 32^o and an angle of 58^o if the bean bag is thrown with an initial velocity of 6 m/s:

                  32^o                                                                   58^o

i.        What are the components of velocity?

            V_x32: Cos 32^o = V_x32/6                                                V_x58: Cos 58^o = V_x58/6           

            6 x Cos 32^o = V_x32 = 5.09 m/s                                    6 x Cos 58^o = V_x58 = 3.18 m/s

            V_y32: Sin 32^o = V_y32/6                                                 V_y58: Sin 58^o = V_y58/6

            6 x Sin 32^o = V_y32 = 3.18 m/s                                     6 x Sin 58^o = V_y58 = 5.09 m/s

ii.      What is the maximum height of the bean bag’s motion?

            t_TOP32 = V_y/10 = 3.18/10 = 0.318s                   t_TOP58 = V_y58/10 = 5.09/10 = 0.509s

            h_MAX32 = ½ x 10 x 0.318² = 0.506 m               h_MAX58 = ½ x 10 x 0.509² = 1.295 m

iii.    How long will the bean bag be in the air?

            t_TOTAL32 = 2 x t_TOP32 = 0.636 s                        t_TOTAL58 = 2 x t_TOP58 = 1.018 s

iv.    How far away from you will the bag land?

            d_x32 = 5.09 x 0.636 =   3.24 m                         d_x58 = 3.18 x 1.018 = 3.24m

v.      If the center of the bull’s-eye ranges from 4.9 m to 5.1 m away from you, does your bean bag win?

                        No                                                                   No

4.      A stunt driver drives a red mustang convertible up a ramp and off a cliff. The car leaves the ramp at a velocity of 60 m/s at an angle of 45^o to the horizontal; the cliff and ramp combined cause the car to begin its projectile motion at a height of 315m above the ground. If you were coordinating this stunt, how far away would you put a landing surface so that your stunt driver was not injured?

In order to find horizontal distance we need horizontal velocity and time. We can find both horizontal and vertical velocity from the initial conditions, but we’ll have to calculate the time it will take for the car to reach the ground. So first we’ll find the components of velocity:

            V_x: Cos 45^o = V_x/60                                       V_y: Sin 45^o = V_y/60

            60 x Cos 45^o = V_x = 42.43 m/s                       60 x Sin 45^o = V_y = 42.43 m/s

Horizontal velocity is constant, but the vertical velocity we calculated above is only the initial vertical velocity.

We use the initial vertical velocity to find the time it takes the car to reach the top of its path and fall to the ground. Let’s think about this in two parts; the time it takes to reach h_MAX first:

t_TOP = 42.43/10 = 4.243 s

Now what about the time it takes to fall from the maximum height? Well first we need to know the maximum height:

h_TOP = ½ x 10 x 4.243² = 90.02 m

h_MAX = h_TOP + h_o =  90.02 + 315 = 405.02 m

Now we calculate the time it takes to fall from a height of 405.02 m:

405.02 = ½ x 10 x t_DOWN²

t_DOWN² = 81.004

t_DOWN = 9.000 s

Putting these two times together, we have the total time it takes the car to travel up to its maximum height and then fall back down. This is the total time in the air and this is the time we will want to use to solve for horizontal distance.

t_TOTAL = t_TOP + t_DOWN = 4.243 + 9.000 = 13.243 s

d_x = 42.43 x 13.243 = 561.9 m

The landing surface is centered 561.9 m from the base of the cliff.

DEFINITION OF SIMPLE HARMONIC MOTION (SHM)

Definition of Simple Harmonic Motion

1. Simple harmonic motion occurs when the force F acting on an object is directly proportional to the displacement x of the object, but in the opposite direction.
   
2. Mathematical statement F = - kx
   
3. The force is called a restoring force because it always acts on the object to return it to its equilibrium position.

An object is undergoing simple harmonic motion (SHM) if;

1. the acceleration of the object is directly proportional to  its displacement from its equilibrium position.
2. the acceleration is always directed towards the equilibrium position.

- See more at: http://physicsnet.co.uk/a-level-physics-as-a2/further-mechanics/simple-harmonic-motion-shm/#sthash.iGsH4mWo.dpuf

Descriptive terms

1. The amplitude A is the maximum displacement from the equilibrium position.
   
2. The period T is the time for one complete oscillation. After time T the motion repeats itself. In general x(t) = x (t + T).
   
3. The frequency f is the number of oscillations per second. The frequency equals the reciprocal of the period.  f = 1/T.
   
4. Although simple harmonic motion is not motion in a circle, it is convenient to use angular frequency by defining w = 2pf = 2p/T.

An object is undergoing simple harmonic motion (SHM) if;

1. the acceleration of the object is directly proportional to  its displacement from its equilibrium position.
2. the acceleration is always directed towards the equilibrium position.

- See more at: http://physicsnet.co.uk/a-level-physics-as-a2/further-mechanics/simple-harmonic-motion-shm/#sthash.iGsH4mWo.dpuf

An object is undergoing simple harmonic motion (SHM) if;

1. the acceleration of the object is directly proportional to  its displacement from its equilibrium position.
2. the acceleration is always directed towards the equilibrium position.

- See more at: http://physicsnet.co.uk/a-level-physics-as-a2/further-mechanics/simple-harmonic-motion-shm/#sthash.72NJe4OB.dpuf

DISTANCE-TIME GRAPH

Interpreting Distance - Time Graph
Generally, time is plotted on the x axis and distance is taken on the y axis of the coordinate system. 

PROJECTILE MOTION GRAPH

Interpreting Projectile Motion Graph

Projectile motion follows the two dimensional kinematics. The trajectory of the projectile is always a parabola. It is not a linear graph.
In the figure below the projectile motion graph is shown.
V_x is showing the Horizontal Component of the velocity of the projectile.
V_y is showing the Vertical Component of the velocity of the projectile.  

 Horizontal Component of the Velocity (V_x)

The horizontal component of the velocity does not changes and the displacements covered by the horizontal components of the velocity are uniform. Horizontal velocity component is equal to the initial velocity component. Gravity force does not affect or does not make any change in the horizontal velocity component of the velocity.

Vertical Component of the Velocity (V_y): 
Vertical component of the velocity does not remain constant during the projectile motion. Gravity force acts on it and changes the vertical component of the velocity of the projectile. The displacements covered by the vertical component of the velocity are not uniform. 

1.During the projectile motion, both the magnitude and direction changes.

2.If the projectile is moving in the upward direction, the vertical component of the velocity is in the upward direction and decrease in its magnitude.
3.When the projectile moves in the downward direction, the direction of the vertical components of the velocity is in the downward direction and increase in the magnitude.

PROJECTILE MOTION EQUATIONS

Projectile Motion Equation is used to find the distance, velocity and time taken in the projectile motion. The Curved path along which the projectile travels is what is known as trajectory. Projectile Motion is the free fall motion of any body in a horizontal path with constant velocity. 

Where V_x is the velocity along x-axis,

       V_xo is the initial velocity along x-axis,
          V_y is the velocity along y-axis,
          V_yo is the initial velocity along y-axis.
  g is the acceleration due to gravity and
         t is the time taken.

 

 Where V_o is the initial Velocity

 sin θ is the component along y-axis,

 cos θ is the component along x-axis.

Solved Examples 

Question 1:  

A body is projected with a velocity of 20 ms^-1 at 50^o to the horizontal. Find
 (i) Maximum height reached
(ii) Time of flight and
(iii) Range of the projectile.
Solution:

Initial Velocity V_o = 20 ms^-1,
θ = 50^o

Time of flight, t = 2V0sinθg
                           = 2×20×sin5009.8
                           = 3.126 s.

Maximum Height reached, H = V2osin2θ2g
                                                    = (20)2sin25022×9.8
                                                    = 11.97 m.

Horizontal Range R = v20sin2θg
                  = 202sin100o9.8
                                    = 40.196 m.

Question 2:
John is on top of the building and jack is down. If john throws a ball at an angle of 60^o and with initial velocity 20 m/s. At what height will the ball reach after 2 s?
Solution: Given: V_yo = 20 m/s,
           Δ t = 2s,
The Vertical velocity in y direction is given by V_y = V_yo sin 60^o
                                                                                     = 20 × 3√2
                                                                                     = 17.32 m/s.
 Vertical distance, y = V_yo t - 12 gt²
                                   = 20 × 2 - 0.5 × 9.8 × 4
                                   = 20.4 m.