2. A projectile is fired with an initial speed of 113 m/s at an angle of 60.0 degrees above the horizontal from the top of a cliff 49.0 m high.

Determine the (a) time to reach maximum height, (b) maximum height above the base of the cliff reached by the projectile, (c) the total time it is in the air, and

(d) horizontal range of the projectile.

Please explain what is the difference between these two questions, and tell me what is the first thing I should look for in a physics question.

In a Physics question I would always look for
the branch of Physics that it applies to – is it to do with momentum, energy, electric current,
density, magnetic fields or what? This will narrow down the area that you have to think
about.

Having sorted that out you may then need to go on to look for a formula if the
question needs a numerical answer. In the formula there will be a lot of symbols but if you
have chosen the correct equation then you will know all the terms in the equation except one
– this is the quantity that you have to work out.

Write down your thoughts as you go,
it is not a good idea to work it all out in your head.

Many of the questions have been
answered in some way before – in spite of what you may think Physics is not a huge subject.
A quote from one of my ex students " Physics is a subject which if you understand it you do
not have to know much". (The student is now a science teacher in the USA).

Now to
the answers to the projectile questions.

The definitions you asked about:

1.
Projectile motion – any motion of an unpowered (strictly) of an object in a gravitational field. If
air resistance is ignored this is a parabolic path.

2. Velocity – distance/time where the
velocity is a vector quantity, this is one which has magnitude as well as direction.

3.
Magnitude – another word for size

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.

We will ignore air resistance for the time being.

Consider the
horizontal motion:

The velocity after time t = v_{x} = u since there is no horizontal
acceleration

Now consider the vertical motion:

The initial vertical velocity (u

v

Therefore v

The velocity (v) after a time t can be found from the equation:

(using pythogoras) and the direction by:

where θ is the angle that the trajectory makes with the horizontal at that point.

A stone is thrown horizontally with an initial velocity of 5ms^{-1} from a
bridge that is 19.6m above the street below.

Calculate:

(a) the time it takes to hit the
ground

(b) the velocity (magnitude and direction) just before it strikes the
street.

(Ignore air resistance in your calculations and take g = 9.8 ms^{-
2}).

(a) Using h = ½ gt^{2} 19.6 = ½ 9.8 x t^{2} and so t = 2 s

(b)
Vertical velocity after 2 s = 0 + gt = 9.8x2 = 19.6 ms^{-1}

Velocity after 2s = [384 +
25]^{1/2} = 20.2 ms^{-1}

Direction of motion to the horizontal tan q= 20.2/5 = 4.04 and so θ =
76^{o}

Consider now the case of an
object that is projected at an angle to the horizontal other than 90^{o}.

It is helpful to treat
the horizontal and vertical components of velocity separately.

Consider an object
projected with velocity u at an angle A to the horizontal.

Vertical component of
velocity = u sin A

Horizontal component of velocity = u cos A

If we ignore the
effects of air resistance, the horizontal velocity is constant and the vertical velocity changes
with a uniform acceleration. The path that the body follows is a parabola as can be seen from
the proof below.

Vertical motion: h = ut sin A – ½ gt^{2}

Horizontal
motion: s = ut cos A

(a) time to reach maximum height is given by the equation v = u + gt where u = 113 sin 60 and v = 0. (At the top of the path the projectile will not be moving upwards).

This gives t = 113sin60/9.8 = 10 s

(b) Vertical velocity on projection from the cliff top will be 113sin60 = 97.9 m/s

maximum height = average velocity x time = ([97.9+0]/2)x10 = 489 m

(c) When the projectile falls back to the level of the top of the cliff it will be travelling downwards at 113 m/s at 60

Horizontal velocity throughout the path = 113cos60 = 56.5 m/s (no horizontal acceleration)

Vertical velocity as it passes the cliff top will be 113sin60 = 97.9 m/s

Vertical velocity at the base of the cliff is given by: v

Time to fall can be found from: v = u + at and so 103 = 97.9 + 9.8t and therefore t = 0.49s

Total time in the air = 20 + 0.49 = 20.5 s

(d) Horizontal range = vt = horizontal velocity x total time = 56.5 x 20.5 = 1158 m