SME
Dynamics · Problem type
Launch speed to reach a target
Use this for “A ball is launched at 45° and must land 40 m away on level ground. How fast must it leave the ground?”
Diagram
vₓ = v₀ cos θ = 19.81 × cos 45°
vₓ = 14 m/s
vᵧ = v₀ sin θ = 19.81 × sin 45°
vᵧ = 14 m/s
H = h + vᵧ²/(2g)
H
=
0
+
14
2
/(2
×
9.81)
=
10
m
In the air 2.86 s, until h + vᵧt − ½gt² = 0.
R = vₓT = 14 × 2.86 = 40 m
This is an example. Type your own values to start a new problem.
R (m)
=
(
v₀ (m/s)
)
2
sin(2
×
θ
°)
÷
g
The picture’s R = vₓT is the same range: vₓ = v₀ cos θ times T = 2v₀ sin θ ÷ g gives v₀² × 2 sin θ cos θ ÷ g = v₀² sin(2θ) ÷ g.
T
Flight time
worked out (not typed)
H
Highest point
worked out (not typed)
Formulas
R
=
v₀
2
×
sin(2 ×
θ
)
9.81
40
=
19.8091
2
×
sin(2 × 45°)
9.81
T
=
2
×
v₀
×
sin(
θ
)
9.81
2.8557
=
2
×
19.8091
×
sin(45°)
9.81
H
=
(
v₀
× sin(
θ
))²
2 × 9.81
10
=
(19.8091 × sin(45°))²
2 × 9.81
Assumptions
•
Level ground: it lands at the height it left from. No air resistance; g = 9.81 m/s² down.
•
Range R = v₀² sin 2θ ÷ g, so v₀ = √(gR ÷ sin 2θ). The longest range for a speed is at 45°.
•
θ and 90° − θ give the same range, so θ is typed, never worked out.
Step-by-step
WE KNOW
R = 40 m, θ = 45°
FIND
launch speed (v₀), flight time (T), highest point (H)
Step 1 · Find launch speed (v₀)
Use R = v₀² × sin(2 × θ) ÷ 9.81
Undo
R
=
v₀
2
sin
2
θ
g
:
times
g,
divide
by
sin
2
θ
,
then
the
square
root.
v₀
=
√(9.81
×
R
sin(2 ×
θ
)
)
v₀
=
√(9.81
×
40
sin(2 × 45°)
)
v₀
=
√(9.81
×
40
sin(90°)
)
v₀
=
√
9.81 × 40
v₀
=
√
392.4
v₀
=
19.8091
m/s
Step 2 · Find flight time (T)
Use T = 2 × v₀ × sin(θ) ÷ 9.81
Gravity
takes
v₀
sin
θ
g
to
stop
the
climb,
and
as
long
to
fall
back.
T
=
2
×
19.8091
×
sin(45°)
9.81
T
=
2
×
19.8091
×
0.70711
9.81
T
=
39.6182
×
0.70711
9.81
T
=
28.014
9.81
T
=
2.8557
s
Step 3 · Find highest point (H)
Use H = (v₀ × sin(θ))² ÷ (2 × 9.81)
At
the
top
the
upward
speed
is
0:
vᵧ
2
=
2gH,
with
vᵧ
=
v₀
sin
θ
at
launch.
H
=
(19.8091 × sin(45°))²
2 × 9.81
H
=
(19.8091 × 0.70711)
2
÷
19.62
H
=
14.007
2
19.62
H
=
196.2
19.62
H
=
10
m
Check
Put the numbers back in. Both sides should match.
40
=
19.8091
2
×
sin(2 × 45°)
9.81
✓
2.8557
=
2
×
19.8091
×
sin(45°)
9.81
✓
10
=
(19.8091 × sin(45°))²
2 × 9.81
✓
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Relative velocity of two moving particles
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