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Projectiles

Table of Contents

Introduction
Projectile Problems
Projectile Answers

       

      

Introduction

We will assume no air resistance in all problems.

Projectiles must be examined as two dimensional vectors.  This motion should be examined as having a horizontal and a vertical vector.

Always break the initial velocity into horizontal and vertical components.  

vo horizontal = vo Cos

   

vo vertical  =  vo Sin

   

     

A key point to remember, there will be acceleration due to gravity in the down direction.   There is NO acceleration in the horizontal direction.

    

There is a special case when the vertical displacement is 0.  This means the projectile ends at the same altitude it begins.

Assign Forward and Down as positive directions.

t = time in the air.  Time is the only variable that is the same for the horizontal and vertical directions.

Since the projectile hits the ground at the same altitude it was shot, x = 0.

vo vertical is negative and a is positive 9.80 m s -2 on Earth.

At its highest point, v vertical = 0
     

                                                                        - vo vertical  
v vertical  =  vo vertical  +  gt  =  0        =>  t = --------------    since either vo or g is negative we can cancel sign
                                                                               g
   

The projectile is in the air twice the time it takes the projectile to reach its highest point.

   
x horizontal = vo horizontal t    (Remember, no acceleration in horizontal direction)

Combining the equations.

  
                     2 vo horizontal  vo vertical          2 vo2  Sin   Cos             vo2 Sin 2 
x horizontal = --------------------------------  =  ----------------------------- = -------------------- = Range
                                     g                                               g                               g                       
    

Sin 90o = 1  Thus the maximum range with no air resistance is achieved when   = 45o.

        
When solved for   
     

                         g Range
           Sin -1 ( -------------- )
                            vo2  
  =  ----------------------------
                        2

      

The above equations can only be used when the projectile lands at the same altitude it was shot.

   

Projectile hits the ground Calculate the time based only on vertical information.  Then use this time to calculate horizontal displacement. 

      

Projectile hits a wall Calculate the time based only on horizontal information.  Then use this time to calculate vertical displacement.

Table of Contents

   

    

Projectile Problems

1. A football is kicked off a tee with an initial velocity of 20.0 m/s with an inclination of 30.0o.  How far downfield does the football hit the ground.
      
2. A soccer ball is kicked with an initial speed of 23.0 m/s trying to hit a spot 50.0 m downfield.  What angle of inclination is necessary?
   
3. An arrow is shot from the top of a 200. m cliff with an initial airspeed of 99.0 m/s and an inclination of 15.0o.  How far from the base of the cliff does the arrow hit the ground.
    
4. A cannon with an inclination of 25.0o is 500. m from a canyon wall.  If the initial airspeed is 150. m/s, how high up the cliff does the cannon ball hit?

Table of Contents

   

   

Projectile Answers

1. A football is kicked off a tee with an initial velocity of 20.0 m/s with an inclination of 30.0o.  How far downfield does the football hit the ground.  How high does the football get above the ground.
      
                 vo2 Sin 2
Range = -----------------
                      g

    

                 (20.0 m s -1)2  Sin (2 x 30.0o)
Range = ---------------------------------------- = 35.3 m
                           9.80 m s -2   

The ball hits at the same altitude it starts from allowing us to use the Range Equation.
 

 

 

 

 

       

vo vertical = vo Sin

    
vo vertical = 20.0 m/s x Sin 30.0o 

   
vo vertical = 10.0 m/s

Calcule the vertical component of the initial velocity.
v = vo  +  at  =  0
    

       - vo        - 10.0 m s -1  
t = ------- = ------------------- = 1.02 seconds
       - a         - 9.80 m s -2    

   
Y = vo vertical t  +  1/2 gt2     

   
Y  =  10.0 m/s x 1.02 s  + 1/2 x 9.80 m s -2 x  (1.02 s)2 

Y  =  5.10 m

The highest point is reached when the vertical component to the velocity is 0.

Up is positive and down is negative.

   

   
Use only the vertical information to find the highest point.

 

    
   
2. A soccer ball is kicked with an initial speed of 23.0 m/s trying to hit a spot 50.0 m downfield.  What angle of inclination is necessary?
   
                                                                 g  Range
                                                     Sin -1 ( -------------- )
                 vo2 Sin 2                                    vo2 
Range = ----------------- => =   ----------------------------  
                      g                                               2

    

                       9.80 m s -2  x 50.0 m
          Sin -1 ( ---------------------------- )
                          (23.0 m s -1 )2 
  =  -------------------------------------------  =  33.9o   
                               2

The ball hits at the same altitude it starts from allowing us to use the Range Equation.
   
   
3. An arrow is shot from the top of a 200. m cliff with an initial airspeed of 99.0 m/s and an inclination of 15.0o.  How far from the base of the cliff does the arrow hit the ground.
    

 

 

 

 

     
      

vo horizontal = vo Cos   
                   = 99.0 m/s x Cos 15.0o 
                   = 95.6 m/s

vo vertical = vo Sin
               = 99.0 m/s x Sin 15.0o 
               = 25.6 m/s

Break the intial velocity into vertical and horizontal components.
Down is positive and up is negative

y = vo vertical t  +  1/2 gt2  

200. m =  - 25.6 m/s t   +  1/2 x 9.8 m/s2  t2     

4.9 m/s2 t2  -  25.6 m/s t -  200. m = 0 

t  =  - 4.29 s   and t  =  9.51 s   (negative time not valid)

   

x = vo horizontal  t            (No Horizontal Acceleration)

x  =  95.6 m/s x 9.51 s  =  909 meters

Calculate the time the arrow is in the air using the vertical information only because the arrow hits the ground.

Solve for t using the quadratic equation.

   

Use the time to calculate the horizontal displacement.

   
    
4. A cannon with an inclination of 25.0o is 500. m from a canyon wall.  If the initial airspeed is 150. m/s, how high up the cliff does the cannon ball hit?
   
 

 

 

 

 

         

vo horizontal = vo Cos   
                   = 150. m/s x Cos 25.0o 
                   = 136 m/s

vo vertical = vo Sin
               = 150. m/s x Sin 25.0o 
               = 63.4 m/s

 

Break the intial velocity into vertical and horizontal components.
x  =  vo horizontal t                   (No Horizontal Acceleration)
    

                  x                500. m
t  =  ----------------  =  ---------------  =  3.68 s
        vo horizontal         136 m s -1   

    

Up is positive and down is negative    

y  =  vo vertical t  +  1/2 gt2    

y  =  63.4 m s -1 x 3.68 s  +  1/2 x ( - 9.80 m s -2 )(3.68 s)2  

y  =  167 m  up the wall

Calculate the time the arrow is in the air using the horizontal information only because the arrow hits the wall of the canyon.

   

Calculate the vertical displacement using the time.

Acceleration is down.

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