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

Acceleration types, units, importance & Graphic representation of moving in a straight line