SME
Grade 11 Science · Problem type
Launched level off a ledge
Use this for “A ball rolls off a 45 m ledge at 8 m/s. How long is it in the air, and how far out does it land?”
Diagram
vₓ = v₀ cos θ = 8 × cos 0° = 8 m/s
vᵧ = v₀ sin θ = 8 × sin 0° = 0 m/s
In the air 3.03 s, until h + vᵧt − ½gt² = 0.
R = vₓT = 8 × 3.03 = 24.2 m
This is an example. Type your own values to start a new problem.
T (s)
=
2
×
h (m)
9.8
m/s²
R (m)
=
v₀ (m/s)
×
T (s)
Formulas
T
=
√
2 ×
h
/9.8
3.03
=
√
2 × 45/9.8
R
=
v₀
×
T
24.2 = 8 × 3.03
Assumptions
•
It leaves level (θ = 0), so it starts with no vertical velocity.
•
It falls as if dropped, while moving across at a steady v₀.
•
No air resistance, and g = 9.8 m/s².
Step-by-step
WE KNOW
v₀ = 8 m/s, h = 45 m
FIND
time in the air (T), range (R)
Step 1 · Find time in the air (T)
Use T = √(2 × h/9.8)
Falling
from
rest
through
h
:
h
=
1
2
gT²,
so
T
=
√
2
h
/g
.
T
=
√
2 × 45/9.8
T
=
√
90/9.8
T
=
√
9.184
T
=
3.03
s
Step 2 · Find range (R)
Use R = v₀ × T
Across,
it
keeps
v₀
for
the
whole
fall.
R
=
8
×
3.03
R
=
24.2
m
Check
Put the numbers back in. Both sides should match.
3.03
≈
√
2 × 45/9.8
✓
24.2 ≈ 8 × 3.03 ✓
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