BAB 4

Bab 4 

Projectile Motion 

Before we continue to this topic, lets see this table first.....

After finish reading this table, let's continue to this topic again!

• Initial speed of the projectile :

            1. Vxo = Vo cos Ө (In horizontal direction)

            2. Vyo = Vo sin Ө (In vertical direction)

•  Speed of the projectile at arbitrary time :

           1. In horizontal direction :

                Vx = Vxo

           2. In vertical direction : 

                Vy = Vyo - gt 

• Position of projectile at arbitrary time :

             1. In horizontal direction : 

                 X = Xo + Vxo t

             2. In vertical direction :

                 y = yo + Vyot - 1/2 gt^2

• Maximum height :

           1.  Time required by the projectile to arrive the peak of its path is :

                                         Tm = Vyo : g 

                     * g = gravity 

            2. The maximum heigth of the path of the projectile is : 

                                          Hm = 1/2 x Vyo^2 : g

• Range and Maximum Range 

            1. Time to reach its path : 

                     T = 2tm

            

            2. Because the horizontal motion is a motion with uniform speed, we find the range of the                       projectile is : 

                    

                    R = VxoT = Vo^2 x sin 2Ө : g

                            

                             Which is equal To....

                  

                    R = Vo^2 x 2 x (cos Ө) x (sin Ө) : g 

Example :

A projectile was shot with initial speed of 100 m/s making an elevation angle of 30。with respect horizontal direction. determine : 

a) Components of initial velocity of the projectile. 

         Vxo = Vo cos Ө = 100 x cos 30 = 100 x √3 /2 = 50√3 m/s

         Vyo = Vo sin Ө = 100 x sin 30 = 100 x 1/2 = 50 m/s

b) Compoents of velocity and position of the projectile one seconds after shooting.

         Vx = Vox = 50√3 m/s

         Vy = Vyo - gt = 50 - 10 x 1 = 40 m/s

    Components of position at t = 1 s

         Xo = 0

         yo = 0

       X = Xo + Vxot = 0 + 50√3 x 1 = 50√3 m

       Y = Yo + Vyot - 1/2 gt^2 = 0 + 50 x 1 - 1/2 x 10 x 1^2 = 45 m 

c) Time required to reach the peak of its path. 

        Tm = Vyo/g = 50/10 = 5 sec

d) The maximu height of the path 

        Hm = 1/2 x Vyo^2/g = 1/2 x 50^2/10 = 125 m

e) The time required by the projectile to reach a height of one fourth its maximum height of the path is 

              h = 1/4 x Hm = 1/4 x 125 = 31.25 m

             

   Use equation 

              h = y - yo = Vyot - 1/2 gt^2

    31.25 = 50 x t - 1/2 x 10 x t^2

    31.25 = 50t - 5t^2 or t^2 - 10t + 6.25 = 0

         T1 = - (-10) - √(-10)^2 - 4 x 1 x 6.25 / 2 x 1 = 10 - 5√3 / 2 s

         T2 = - (-10) - √(-10)^2 - 4 x 1 x 6.25 / 2 = 10 + 5√3 / 2 s

 From this calculation, we found moments in time when the projectile reached one fourth of if maximum height. The smaller time is required when the projectile is moving up and the longer one is the time required when the projectile is moving back down to the ground. 

f) the time required to reach the ground again :

        T = 2Tm = 2 x 5 = 10 sec

g) The range of the projectile : 

        R = VyoT = 50 x 10 = 500 m