SME
Grade 11 Science · Physics
Motion in two dimensions: vectors and projectile motion
Use this for “A ball is launched at 20 m/s, 30° above level, from 1.5 m up. How long is it in the air, how far does it go, and how high?”
Diagram
vₓ = v₀ cos θ = 20 × cos 30°
vₓ = 17.3 m/s
vᵧ = v₀ sin θ = 20 × sin 30°
vᵧ = 10 m/s
H
=
h
+
vᵧ
2
/(2g)
H
=
1.5
+
10
2
/(2
×
9.8)
=
6.6
m
In the air 2.18 s, until h + vᵧₜ − ½gt² = 0.
R = vₓT = 17.3 × 2.18 = 37.8 m
This is an example. Type your own values to start a new problem.
v₀
Launch speed
example
θ
Launch angle
example · °
h
Launch height
example
vₓ
Horizontal velocity
worked out (not typed)
vᵧ
Vertical launch velocity
worked out (not typed)
T
Time in the air
calculated
R
Range
calculated
H
Maximum height
calculated
Formulas
vₓ
=
v₀
×
cos(
θ
)
17.32 = 20 × cos(30°)
vᵧ
=
v₀
×
sin(
θ
)
10 = 20 × sin(30°)
T
=
(vᵧ
+
√
vᵧ
2
+ 2 × 9.8 ×
h
)/9.8
2.181
=
(10
+
√
10
2
+ 2 × 9.8 × 1.5
)/9.8
R
=
vₓ
×
T
37.8 = 17.32 × 2.181
H
=
h
+
vᵧ
2
/(2
×
9.8)
6.6
=
1.5
+
10
2
/(2
×
9.8)
Assumptions
•
No air resistance, and g = 9.8 m/s² downward.
•
Across nothing pushes, so vₓ stays the same; vᵧ drops by 9.8 m/s every second.
•
It lands on level ground h below the launch point (h = 0 for a launch from the ground).
Step-by-step
WE KNOW
v₀ = 20 m/s, θ = 30°, h = 1.5 m
FIND
horizontal velocity (vₓ), vertical launch velocity (vᵧ), time in the air (T), range (R), maximum height (H)
Step 1 · Find horizontal velocity (vₓ)
Use vₓ = v₀ × cos(θ)
The
horizontal
part
of
the
launch
velocity:
v₀
times
cos
θ
.
vₓ = 20 × cos(30°)
vₓ = 20 × 0.866
vₓ = 17.32 m/s
Step 2 · Find vertical launch velocity (vᵧ)
Use vᵧ = v₀ × sin(θ)
The
vertical
part
of
the
launch
velocity:
v₀
times
sin
θ
.
vᵧ = 20 × sin(30°)
vᵧ = 20 × 0.5
vᵧ = 10 m/s
Step 3 · Find time in the air (T)
Use T = (vᵧ + √(vᵧ² + 2 × 9.8 × h))/9.8
It
lands
when
h
+
vᵧ
t
−
1
2
gt²
=
0:
the
positive
root
of
the
quadratic.
T
=
(10
+
√
10
2
+ 2 × 9.8 × 1.5
)/9.8
T
=
(10
+
√
100 + 2 × 9.8 × 1.5
)/9.8
T
=
(10
+
√
100 + 19.6 × 1.5
)/9.8
T
=
(10
+
√
100 + 29.4
)/9.8
T
=
(10
+
√
129.4
)/9.8
T
=
(10
+
11.375)/9.8
T
=
21.375/9.8
T
=
2.181
s
Step 4 · Find range (R)
Use R = vₓ × T
Across, the speed stays vₓ for the whole flight.
R
=
17.32
×
2.181
R
=
37.8
m
Step 5 · Find maximum height (H)
Use H = h + vᵧ²/(2 × 9.8)
At
the
top
vᵧ
is
0:
the
height
gained
is
vᵧ
2
over
2g.
H
=
1.5
+
10
2
/(2
×
9.8)
H
=
1.5
+
10
2
/19.6
H
=
1.5
+
100/19.6
H
=
1.5
+
5.102
H
=
6.6
m
Check
Put the numbers back in. Both sides should match.
17.32 ≈ 20 × cos(30°) ✓
10 = 20 × sin(30°) ✓
2.181
≈
(10
+
√
10
2
+ 2 × 9.8 × 1.5
)/9.8
✓
37.8 ≈ 17.32 × 2.181 ✓
6.6
≈
1.5
+
10
2
/(2
×
9.8)
✓
Topic check
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