Projectiles and answering Physics questions
Question:
1. A stone is thrown horizontally outward from the top of a bridge. The stone is released 19.6
meters above the street below. The initial velocity of the stone is 5.0 m/s. Determine the (a)
total time that the stone is in the air and (b) magnitude and direction of the velocity of the
projectile “just before it strikes the street.
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.
Answer:
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
Object projected horizontally
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 = vx = u since there is no horizontal acceleration
The horizontal distance travelled (s) = horizontal velocity x time = vxt = ut
(1)
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Now consider the vertical motion:
The initial vertical velocity (uy) = 0 and so the vertical velocity after a time t is given by
vy = uy + gt
Therefore vy = gt
The vertical distance travelled (h) = uyt + ½ (gt2) = ½ (gt2) since uy = 0.
(2)
The velocity (v) after a time t can be found from the equation:
Velocity after a time t
v = (vx2 + vy2)1/2
(using pythogoras) and the direction by:
Direction of motion after time t
tan = vy/vx
where  is the angle that the trajectory makes with the horizontal at that point.
Answer to 1
A stone is thrown horizontally with an initial velocity of 5 ms -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 gt2 19.6 = ½ 9.8 x t2 and so t = 2 s
(b) Vertical velocity after 2 s = 0 + gt = 9.8x2 = 19.6 ms -1
Velocity after 2s = √[384 + 25] = 20.2 ms-1
Direction of motion to the horizontal tan  = 20.2/5 = 4.04 and so  = 76o
Object projected at an angle
Consider now the case of an object that is projected at an angle to the horizontal other than
90o.
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 gt2
Horizontal motion: s = ut cos A
Answer 2
Initial speed = 113 m/s at an angle of 60o to the horizontal from a cliff of height 49.0 m
(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
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(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 60o to the horizontal and will have been moving for 20 s.
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: v2 = u2 + 2as = 97.92 + 2x9.8x49 so v =
103 m/s
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
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