SME
Grade 11 Science · Problem type
Free fall from rest
Use this for “A stone falls from rest for 3 s. How far does it fall, and how fast is it going?”
Chart
v = v₀ + at = 0 + (−9.8) × 3
v = −29.4 m/s
Δx
=
v₀ + v
2
×
t
Δx
=
0 + (−29.4)
2
×
3
=
−44.1
m
The shaded area below the axis is a displacement backward (−).
Drop: 44.1 m
This is an example. Type your own values to start a new problem.
d (m)
=
½
×
9.8
m/s²
×
(
t (s)
)
2
The picture’s displacement Δx is the drop d with a minus sign: Δx = −d.
v
Velocity
worked out (not typed)
Formulas
v
=
−9.8
×
t
−29.4 = −9.8 × 3
d
=
1
2
×
9.8
×
t
2
44.1
=
1
2
×
9.8
×
3
2
Assumptions
•
Dropped from rest, so it starts at v₀ = 0.
•
Air resistance is ignored: every object falls with a = −9.8 m/s², whatever its mass.
•
Up is +, so the velocity is negative on the way down; the drop d is how far it fell.
Step-by-step
WE KNOW
t = 3 s
FIND
velocity (v), drop (d)
Step 1 · Find velocity (v)
Use v = −9.8 × t
From rest, the velocity is g times the time, downward.
v
=
−9.8
×
3
v
=
−29.4
m/s
Step 2 · Find drop (d)
Use d = ½ × 9.8 × t²
From
rest
the
drop
is
1
2
gt²:
it
grows
with
the
square
of
the
time.
d
=
1
2
×
9.8
×
3
2
d
=
1
2
×
9.8
×
9
d
=
4.9
×
9
d
=
44.1
m
Check
Put the numbers back in. Both sides should match.
−29.4 = −9.8 × 3 ✓
44.1
=
1
2
×
9.8
×
3
2
✓
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