1. ANSWERS
Question 1
(a) A rock which is simply dropped off a bridge is not an example of a projectile
because whilst the only force acting on it is gravity (like a projectile), there is no
horizontal component to its velocity and the path of the rock is not parabolic in
shape.
(b) The rock could become a projectile if it were launched horizontally from the
bridge.
(c) The rock falls with constant acceleration so we would use a kinematic equation to
solve this problem:
+ d=? d = vit + 1at2 but vi = 0 so vit = 0
vi = 0 2
-
vf =
=> d = 1at2 = 1 x -10 x 42 = -80 m
a = -10
2 2
t =4
This means that the
displacement is -160 m. It is
160 m from the bridge in the
downward direction
Sunday, 25 April 2010

2. (d) The rock is now a projectile so we need to accept that there is also now a
horizontal component to its velocity. The horizontal component is independent of
the vertical component which means that even though the rock is travelling
horizontally this does not change the rate at which it is falling vertically so the rock
will still take 4 s to reach the water.
(e) Very simple solution: d = v.t = 6 x 4 = 24 m
Question 2
(a) (b)
10 ms-2
20 N 20 N
a 10 ms-2
20 N
10 ms-2 10 ms-2 20 N
Fw = mg = 2 x 10 = 20 N
Sunday, 25 April 2010

3. Question 3
(a)
200 ms-1
Path of plane
(b) We need to calculate the time the
plane takes to travel a distance of 3
km horizontally at a horizontal velocity
of 200 ms-1
t = d = 3000 = 15 s
v 200 Path of nut
(c) The altitude of the plane is given by the vertical displacement of the nut. We
need to consider the vertical information:
+ d=? d = vit + 1at2 but vi = 0 so vit = 0
- vi = 0 2
vf =
a = -10 => d = 1at2 = 1 x -10 x 152 = -1125 m
t = 15 2 2
Sunday, 25 April 2010

4. Question 4
21 ms-1
Vertical component, vy
40o
Horizontal component, vx
(a) vx = 21cos40o = 16.09 ms-1
(b) vy = 21sin40o = 13.5 ms-1
(c) At the top of the ball’s path, the vertical component of its velocity is 0 ms-1. The
ball, however is still moving horizontally with a constant speed of 16.09 ms-1.
The instantaneous velocity is therefore given by the following vector:
16.09 ms-1
(d) The vertical height of the ball at the top is given by the displacement when the
vertical velocity is 0 ms-1:
+ d=? vf2 = vi2 + 2ad => d = vf2 - vi2
vi = 13.5 2a
-
vf = 0
a = -10 d = 02 - 13.52 = 9.1 m
2 x -10
t =
Sunday, 25 April 2010

5. (e) We are being asked the time taken for the ball to reach the maximum height
(9.1 m).
+ d = 9.1 vf = vi + at => t = vf -vi
vi = 13.5 a
-
vf = 0 = 0 - 13.5
-10
a = -10
= 1.35 s
t =
(f) In order to calculate the range, d of the ball, we require the size of the
horizontal velocity and the time for which the ball is travelling at this speed. The
range is covered in twice the time it takes for the ball to reach its highest point:
t = 2 x 1.35 = 2.7
v = 13.5 d = v.t = 16.1 x 2.7 = 43.5 m
Sunday, 25 April 2010

6. Question 5
30
30sin60o = 26
60o 25
30cos60o = 15
60
To determine whether the potato hits the tree or not we need to know the vertical
displacement of the potato when it has covered a horizontal distance of 60 m:
V H The time the potato takes to travel a horizontal distance
+
of 60 m can be calculated by analysis of the horizontal
d=? d = 60
- motion and used in the analysis of vertical motion to
vi = 26 v = 15 calculate the vertical displacement of the potato.
vf = t =?
a = -10
Horizontal motion: t = d = 60 = 4 s
t =? v 15
Vertical motion: d = vit + 1at2
2
=> d = 26 x 4 + 1 x -10 x 42
2
= 24 m Because the tree is 25 m high and the vertical
displacement is 25 m then the potato hits the
tree
Sunday, 25 April 2010

7. Question 6
In this problem we have information relating to the maximum height of the ball
which occurs at time equal to half the time required for the ball to complete its
journey.
25 m
100 m
Let t = the time taken for the ball to reach its maximum height.
+ V H In order to calculate the initial velocity of the ball we
will need to calculate the initial vertical velocity of the
d = 25 d = 50
- ball,vi and the horizontal velocity of the ball, v.
vi = ? v=?
(both ?)
t=?
vf = 0
(i) Calculating the initial vertical velocity:
a = -10
vf2 = vi2 + 2ad => vi2 = vf2 - 2ad
t=?
= 02 - 2 x -10 x 25
= 500
=> v = 500
i
= 22.4
Sunday, 25 April 2010

8. (ii) In order to calculate the initial horizontal velocity we need to find t (the time
taken for the ball to travel 100 m horizontally. Calculating t can be done using
information about vertical motion:
vf = vi + at => t = vf - vi
a
= 0 - 22.4
- 10
= 2.24 (time for half the flight) => 2 x 2.24 = time for
Now we can calculate horizontal velocity: the entire flight.
v = d = 100 = 22.32 Remember that since horizontal velocity is
t 4.48 constant, this is the initial horizontal velocity.
We now have the two components of the initial velocity and can draw a vector
diagram to calculate the initial velocity:
vi = 22.322 + 22.42
vi 22.4 ms-1 = 31.6 ms-1
22.32 ms-1
Sunday, 25 April 2010

9. Question 7
(a) If the javelin behaves like an ideal projectile then the shape of its path will be
parabolic.
Where Fw = the force due to gravity (or the weight
(b)
force)
Fw
(c) Let vx = the horizontal component of the ball’s
velocity
30
vx = 30cos 40o
40o = 23 ms-1
vx
Sunday, 25 April 2010

10. (d) The range is the horizontal distance travelled by the projectile. We now know the
horizontal speed but will need the time if we are to calculate the range. We will
need to consider both vertical and horizontal information:
Note:
V H
(i) For the vertical information we can only get
+ d= d=? the 3 bits of information required when we
vi = 30sin40o v = 23 consider half the flight path because at the
- t=? highest point we know that vf = 0
= 19.3
vf = 0 (ii) the initial vertical velocity is the vertical
a = -10 component of the initial velocity
t=?
Calculating the time for half the flight:
vf = vi + at t = vf - vi
a
= 0 - 19.3
- 10 So the time for the entire flight
= 2 x 1.93 = 3.86 s
= 1.93
Now we can calculate the range: d = v.t = 23 x 3.86 = 88.8 m
Sunday, 25 April 2010