SME
Grade 12 Math · Problem type
A launch as parametric equations
Use this for “A ball is thrown at 20 m/s, 30° up. Where is it after 1 s, and when does it land?”
Diagram
vₓ = v cos θ = 20 × cos 30°
vₓ = 17.3 m/s
vᵧ = v sin θ = 20 × sin 30°
vᵧ = 10 m/s
x(t) = 17.3t = 17.3 × 1 = 17.3 m
y(t) = 1 + 10t − 4.9t² = 6.1 m
This is an example. Type your own values to start a new problem.
v
Launch speed
example
θ
Launch angle
example · °
h
Launch height
example
t
Time
example
x
Distance across at t
calculated
y
Height at t
calculated
T
Time to land
worked out (not typed)
Formulas
x
=
v
×
cos(
θ
°)
×
t
17.3205 = 20 × cos(30°) × 1
y
=
h
+
v
×
sin(
θ
°)
×
t
−
4.9
×
t
2
6.1
=
1
+
20
×
sin(30°)
×
1
−
4.9
×
1
2
0
=
h
+
v
×
sin(
θ
°)
×
T
−
4.9
×
T
2
0
=
1
+
20
×
sin(30°)
×
2.1363
−
4.9
×
2.1363
2
Assumptions
•
No air resistance; g = 9.8 m/s², so the fall is 4.9t² metres.
•
x = (v cos θ)t and y = h + (v sin θ)t − 4.9t²: the parameter t is the time.
•
It lands when y = 0: the positive root of that quadratic in t.
Step-by-step
WE KNOW
t = 1 s, v = 20 m/s, θ = 30°, h = 1 m
FIND
distance across at t (x), height at t (y), time to land (T)
Step 1 · Find distance across at t (x)
Use x = v × cos(θ°) × t
Across,
the
speed
stays
v
cos
θ
the
whole
flight.
x
=
20
×
cos(30°)
×
1
x
=
20
×
0.866
x
=
17.3205
m
Step 2 · Find height at t (y)
Use y = h + v × sin(θ°) × t − 4.9 × t²
Start
at
h
,
rise
at
v
sin
θ
,
and
fall
4.9
t
².
y
=
1
+
20
×
sin(30°)
×
1
−
4.9
×
1
2
y
=
1
+
20
×
0.5
−
4.9
y
=
1
+
10
−
4.9
y
=
6.1
m
Step 3 · Find time to land (T)
Use 0 = h + v × sin(θ°) × T − 4.9 × T²
Set
y
=
0:
−
4.9
T
²
+
10
T
+
1
=
0,
so
a
=
−4.9,
b
=
v
sin
θ
=
10
and
c
=
h
=
1.
The
quadratic
formula’s
positive
root
is
when
it
lands.
T
=
(
v
×
sin(
θ
°)
+
√((
v
×
sin(
θ
°))²
+
19.6
×
h
))
÷
9.8
T
=
(20
×
sin(30°)
+
√((20
×
sin(30°))²
+
19.6
×
1))
÷
9.8
T
=
10 +
√
10
2
+ 19.6 × 1
9.8
T
=
10 +
√
119.6
9.8
T
=
10 + 10.9362
9.8
T
=
20.9362
9.8
T
=
2.1363
s
Check
Put the numbers back in. Both sides should match.
17.3205 ≈ 20 × cos(30°) × 1 ✓
6.1
=
1
+
20
×
sin(30°)
×
1
−
4.9
×
1
2
✓
0
≈
1
+
20
×
sin(30°)
×
2.1363
−
4.9
×
2.1363
2
✓
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