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Projectiles vertical motion

A body that is projected through a gravitational field is known as a projectile. We will look for the present at the case of an un-powered projectile such as a stone and ignore the effects of air resistance.

The vertical acceleration is due to the gravitational attraction of the Earth and is called the acceleration due to gravity (g).

Above the Earth's surface the value of g decreases as the distance from the centre of the Earth increases, but if we restrict our considerations to points close to the Earth's surface its value is sensibly constant and has a value of about 9.81 ms- 2 (often simplified to 10 ms-2)

Since this acceleration is produced by the gravitational field of the Earth it may also be called the gravitational intensity (units Nkg-1). The value of g can vary due to the following factors:
(a) distance above sea level - g is less up a mountain 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-spherical 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.


Object projected vertically upwards

We will look first at the case of an object that is projected vertically upwards.
Initial vertical velocity = u
We will call time of flight 2t, (in other words the time to return to the ground again) and so the time to reach the top is t. You should understand that it will take the same time for the projectile to reach the top of its path as it will to return to the ground again.
Velocity at the top of the path (v) = 0, it is not moving at this point.

1. Maximum height reached (h)
Since the time to reach the maximum height is the same for an object released from rest at that height to fall to the ground we can use:

Using h = ½ gt2
Alternatively if we know the initial velocity.
v2 = u2 + 2as with s = h and v = 0 at the top of the path and a = -g then:

h = u2/2g

notice the negative sign because the object is slowing down as it rises, g acts in the opposite direction to the velocity of the projectile. Also remember that the acceleration of the projectile is 9.8 ms-2 towards the ground throughout the trajectory, even at the very top.


Example problem
A ball is thrown vertically upwards with an initial velocity at 30 ms-1
Calculate:
(a) the maximum height reached
(b) the time taken for it to return to the ground
(a) Using v2 - u2 + 2as
0 = 900 - 2x10xs 20s= 900 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.
(b) Using v = u + at, 30 = -30 + 10t t = 6s
Remember that the ball must return to the ground with the same speed with which it left it.

2. Any other height than the maximum

h = ut + ½ gt2
v2 = u2 + 2gh

where as before when the numbers are put into the equation g = - 9.8 ms- 2.

Example problem
A balloon is going up at a steady velocity of 12 ms-1 at a height of 32 m above the ground when an object is dropped from it.
How long does it take for the object to reach the ground?

Consider first the time taken to return to the 32 m level.

v = 12ms-1 and u = -12ms-1
(notice the minus sign that allows for the different direction of u and v. It does not matter which one you choose to be minus as long as g is also made plus or minus).
Therefore taking g = 10 ms-2
Using: v = u+at spacewe have: 12 = - 12 + 10t
Therefore t = 2.4 s.

Now it falls to the ground.
Using s = ut + ½at2 space 32 = 12t+ 5t2
This is solved to give t = 1.65 or - 4.00s.
The negative value indicates a time before it passed the 32 m level, which is clearly of no physical significance here.
Therefore the total time = 1.6 s +2.45 s = 4s


 
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© Keith Gibbs 2013