# Safety skills , Applications of motion with uniform acceleration ( Free fall & Projectiles )

**Safety skills**

**To avoid the dangers of exceeding prescribed speeds and to save souls , Traffic instructions should be followed such as leaving an appropriate distance between vehicles to allow the driver to stop safely in case of emergency , ****Obviously, More spacing between vehicles is required as Speed of cars gets higher , The road is wet or has oil stains , Trucks should leave larger spacing than small cars .**

**Free fall **

**If two objects of different masses ( a book and a sheet of paper ) are dropped at the same time from a high point , the two objects start motion from rest ( v _{i} = 0 ) falling under the effect of two forces :**

**The gravitational pull ( their weights ) and the air resistance , Since collision of the object with air molecules affects the velocity of falling of light objects ( the paper sheet ) more than that of heavier objects ( the book ) , we find that the book reaches the ground first .**

**If the air resistance is neglected , The two objects fall under the effect of their weights only and acquire a uniform acceleration that acts to increase the speed of falling gradually till it reaches its maximum value when touching the ground , this acceleration is called acceleration due to gravity or free fall acceleration .**

**Galileo proved that Falling objects of different masses , when neglecting air resistance reach the ground at the same time , He put an end for Aristotle’s idea that implied , ” Heavy objects would reach the ground in a shorter time than that taken by lighter objects ” , ****He proved this by dropping two objects of different masses down Tower of Pisa in Italy .**

**Free fall acceleration ( g )**

**It is the uniform acceleration by which objects move during free fall towards the ground , This acceleration varies slightly from one position to another depending on its distance from the Earth’s center , Its average value equals 9.8 m / s² and for simplicity , it can be considered 10 m / s² .**

**When free fall acceleration of an object = 9.8 m / s² , This means that the velocity of the object that falls freely increases by 9.8 m / s every second .**

**When an object falls freely downwards , The velocity of object increases gradually till it reaches its maximum value when reaching the ground , Its initial velocity v _{i} = 0 .
**

**g = Δ v / Δ t = ( v _{f} − v_{i} ) / ( t − 0 ) = v_{f} / t **

**The free fall acceleration ( g ) is positive ( increasing velocity ) since the direction of object motion is in the same direction of Earth’s gravity .**

**When an object is projected vertically upwards , The velocity of object decreases gradually till it vanishes at the maximum height , Its final velocity v _{f} = 0 .**

**g = Δ v / Δ t = ( v _{f} − v_{i} ) / ( t − 0 ) = − v_{i} / t**

**The free fall acceleration ( g ) is negative ( decreasing velocity ) since the direction of object motion opposes the direction of Earth’s gravity .**

**Projectiles**

**Vertical projectiles**

**When an object is projected vertically upwards , it starts at initial velocity ( v**_{i}) which does not equal zero with uniform deceleration ( − 10 m / s² ) .**Velocity of object decreases gradually as the object gets higher and reaches zero at maximum height .****Direction of velocity changes when the object returns back to the ground under the effect of the Earth’s gravity that makes the object accelerate ( 10 m/s² ) .****Velocity of the object when projected up = − Its velocity at the same point on falling .****Time of rising = Time of falling .**

**Projectiles projected at angle ( Motion in two dimensions )**

**When a ball is projected upwards at an angle ( θ ) to the horizontal , It moves in a curved path , we can resolve velocity in two dimensions , Horizontal ( x ) and vertical ( y ) as shown :**

**In the horizontal dimension ( x ) , **** The ball velocity is uniform ( v _{ix} ) ( neglecting any friction ) : **

** v _{ix} **=

**v**

_{i}_{ }cos θ**Substituting the value of ( v _{ix} ) in the three equations of motion where : a_{x }= 0 , then : v_{fx} = v_{ix} **

**In the vertical dimension ( y ) , **** The ball moves at the acceleration due to gravity , Consequently , velocity varies , the initial velocity in the vertical dimension ( v _{iy} ) is found by the relation :**

** v _{iy} = v_{i } sin θ**

**Substituting the value of ( v _{iy} ) in the three equations of motion by considering ( a_{y }= − 10 m / s² ) .**

**The velocity of the projectile at any instant is given by Pythagoras’ relation :**

**(v _{f} )² **=

**(v**+

_{fx})²**(v**

_{fy})²**Finding the time of reaching the maximum height ( t ) :**

**Substituting by v _{fy} = 0 in the first equation of motion , we find :**

**0 = v _{iy} + g t , t = − v_{iy} / g**

**Time taken till returning back to the plane of projection ( flight time ) : **

**T = 2 t = ( − 2 v _{iy} ) / g**

**Finding the maximum height reached by the projectile ( h ) :**

**Substituting by v _{fy} = 0 in the third equation of motion , we find :
**

**2 g h = − ( v _{iy} )² , h = **

**− ( v**

_{iy})² / 2 g**Finding the horizontal range ( the horizontal distance reached by the projectile ) ( R ) : **

**Time of the maximum horizontal range = Flight time = T **

**Substituting by ( a _{x }= 0 ) and ( d = R ) in the second equation of motion , we find :**

**R = v _{ix} T = 2 v_{ix} t = ( − 2 v_{ix} v_{iy} ) / g**

**The projectile reaches maximum horizontal range when it is projected at an angle 45° , ****The horizontal range is the same when the projectile at complementary angles ( Angles of sum 90° ) .**

**Guidelines to solve problems**

**Equations of motion are not applied in x – axis :**

**Because a _{x }= 0 , v_{ix} = v_{fx }, then : v = d / t ⇒ v_{x }= R / T**

**Equations of motion are not applied in y – axis :**

**v _{f} = v_{i} + at ⇒ v_{fy} = v_{iy} + gt**

**d = v _{i} t + ½ a t² ⇒ h = v_{iy} t + ½ g t²**

**(v _{f} )² **=

**(v**+

_{i})²**2 a d ⇒**

**(v**=

_{fy})²**(v**+

_{iy})²**2 g h**

**(v _{i} )² **=

**(v**+

_{ix})²**(v**

_{iy})²**(v _{f} )² **=

**(v**+

_{fx})²**(v**

_{fy})² , d² = h² + R²
## Recent Comments