# Free fall An object undergoing free fall has an acceleration of 9.8118m/s2 . This varies from the equator when it is 9.7803m/s2 to the poles when it.

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Free fall An object undergoing free fall has an acceleration of m/s2 . This varies from the equator when it is m/s2 to the poles when it is m/s2

You can solve π =π’π‘+ 1 2 π π‘ 2 using the quadratic formula
You can solve π =π’π‘+ 1 2 π π‘ 2 using the quadratic formula. You just need to set the equation equal to 0. 1 2 π π‘ 2 +π’π‘ βπ =0 Then use the quadratic formula to find t. Here: x = t, a = a, b = u, c = s Remember, if you end up with a time less than 0, to reason that this solution is unphysical, and then just accept the remaining solution

Factors affecting g Β The value of g can vary due to the following factors: (a) distance above sea level - g is less up a mounΒ­tain or in a satellite or plane; (b) on the equator - the distance from the centre of the Earth is greater here, because of the non-spherΒ­ ical shape of the Earth, and g is smaller; (c) above mineral deposits there is a greater attraction and so g is larger; (d) the centripetal force of the rotating Earth causes a variation in g from Equator to pole.

Acceleration due to gravity and gravitational field strength
Weight = mass x gravitational field strength. But according to Newtonβs second law Force = mass x acceleration. The force = weight in this situation Therefore numerically Gravitational field strength (N/kg) = acceleration due to gravity. So on earth the acceleration due to gravity = 9.81m/s2 near the earthβs surface

Total time for up and down=2x30=60secs
Example 1 A gun is fired vertically into the air. If the bullet travels at 300m/s calculate how long it takes for the bullet to land and how high it travels? u=300m/s v=0 s=? a=-10m/s2 t=? At top of flight Use v2=u2+2as 0=90,000+(2x-10xs) 0=90,000-20s 20s=90, s=90,000/20=4500m v=u+at 0=300-10t 10t= t=30secs Total time for up and down=2x30=60secs

Example problem 1 A ball is thrown vertically upwards with an initial velocity at 30 ms-1 Calculate: (a) the maximum height reached, (a) Using v2 = u2 + 2as, 0 = x10xs s= s =45m Notice that at the maximum height the vertical velocity is zero and that the acceleration due to gravity is negative since it acts to retard the ball.

Example 1 continued (b) the time taken for it to return to the ground.
(b) Using v = u + at, = t t = 6s Remember that the ball must return to the ground with the same speed with which it left it.

Object projected horizontally
We will now look at an object projected horizontally from a cliff with a velocity u. The problem is to find out where this object will be after a time t and how fast it will be travelling and in what direction. Β

The important thing to remember is that you can consider the motion in two parts :-
(a) motion in the horizontal direction - this is uniform velocity since no forces act in this direction (b) motion in the vertical direction - this is uniformly accelerated motion due to the gravitational pull of the Earth, the vertical acceleration being the strength of the Earth's field (g = 9.8 ms-2). Remember that this always acts vertically downwards.

The horizontal and vertical motions are completely independent of each other.
We will ignore air resistance for the time being. Β  Consider the horizontal motion: The horizontal distance travelled (s) = horizontal velocity x time = vxt = ut (1)

Now consider the vertical motion:
The initial vertical velocity (uy) = 0 So the vertical velocity after a time t is given by vy = uy + gt Therefore vy = gt The vertical distance travelled (h) = uyt + Β½ (gt2) = Β½ (gt2) (2) since uy = 0.

Finally Velocity after a time t v = (vx2 + vy2)1/2
The velocity (v) after a time t can be found from the equation: Β  Β Direction of motion after time t tanπ­ = vy/vx Β (using pythogoras) and the direction by: Β  Β where π­ is the angle that the trajectory makes with the horizontal at that point

Example A ball is thrown horizontally with an initial velocity of 6 ms-1 from an open window that is 4m above the ground. Calculate: (a) the time it takes to hit the ground (b) the distance from the wall where it hits the ground (c) the velocity (magnitude and direction 0.5 seconds after it is thrown. (Ignore air resistance in your calculations and take g = 9.8 ms-2).

Example 2 A ball is kicked off a cliff with a speed of 10m/s. If the cliff is 100m above the sea, calculate how far from the base of the cliff the ball lands.

Projectiles A Parabola

Projectiles - horizontal direction
Constant speed No forces act A Parabola

Constant acceleration
Projectiles Vertical direction Only gravity acts Constant acceleration A Parabola

Projectiles A Parabola

How to solve projectile problems
1) Separate vertical and horizontal parts of motion 2) For vertical motion use the kinematics equations 3) For horizontal motion use dst.

Vertical component of velocity = usinπ­
Horizontal component of velocity = u cos π­ π­π­ π­π­ π­π­

Horizontal motion: s = u cos π­t
If we ignore the effects of air resistance, the horizontal velocity is constant and the vertical velocity changes with a uniform acceleration. Β Vertical motion: h = u sin π­tβ Β½ gt2 Horizontal motion: s = u cos π­t π­π­ π­π­ π­π­

How to solve projectile problems
1) Separate vertical and horizontal parts of motion 2) For vertical motion use the kinematics equations 3) Use s=ut+1/2at2 4) Vertically u=0 so use s=1/2at2 5) Solve to find the time taken to hit the ground 5) For horizontal motion use d=sxt.

How to solve projectile problems
A ball is kicked off a 100m tall cliff and lands 300m from the base. How fast was it kicked off? A ball is kicked vertically into the air and reaches the top of a 15m building. How fast was it kicked and how long did it spend in the air? A bowler bowls a ball from 2.3m above the ground at a speed of 20m/s. Calculate the balls velocity just as it hits the ground. Another ball is thrown from 2 m above the ground if it is to hit stumps 1m tall 10ms away how fast must the ball be bowled?

Projectiles and air resistance
Objects moving through air are slowed down due to air resistance, sometimes called drag. This air resistance affects a spacecraft when it re- enters the Earthβs atmosphere but also the path of a projectile such as a bullet or a ball. When air resistance is taken into account the trajectory of a projectile is changed. The resistance is often taken as being proportional to either the velocity of the object or the square of the velocity of the object. Β The medieval scientists believed that a projectile went upwards at an angle along a straight path, then went through a short curved section before falling vertically back to the ground again. Β Both the range of a projectile and the maximum height that it reaches are affected by air resistance. The mathematics of the motion is quite complicated (especially if you consider the change in the shape and/or surface of a projectile and the variation of the density of the air with height) but the following diagrams try to simplify things by showing generally how air resistance affects both the trajectory and the velocity of a projectile.

Β  The blue lines show the projectile with no air resistance and the red lines show what happens when air resistance is taken into account. The maximum height, the range and the velocity of the projectile are all reduced.

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