SME
University Physics I: Mechanics · Problem type
A projectile’s position and velocity at a time
Use this for “A ball is thrown from 2 m up at 20 m/s, 60° above level. Where is it 2.5 s later, and how fast and which way is it moving?”
Diagram
vₓ = v₀ cos θ = 20 × cos 60°
vₓ = 10 m/s
vᵧ = v₀ sin θ = 20 × sin 60°
vᵧ = 17.3 m/s
x(t) = 10t = 10 × 2.5 = 25 m
y(t) = 2 + 17.3t − 4.905t²
y(t) = 14.6 m
This is an example. Type your own values to start a new problem.
x (m)
=
v₀ (m/s)
cos(
θ
°)
×
t (s)
y (m)
=
h (m)
+
v₀ (m/s)
sin(
θ
°)
×
t (s)
−
½(9.81)
(
t (s)
)
2
vₓ
Horizontal velocity
worked out (not typed)
vᵧ
Vertical velocity at t
worked out (not typed)
v
Speed at t
worked out (not typed)
φ
Direction at t
worked out (not typed) · °
Formulas
vₓ
=
v₀
×
cos(
θ
°)
10 = 20 × cos(60°)
vᵧ
=
v₀
×
sin(
θ
°)
−
9.81
×
t
−7.2045 = 20 × sin(60°) − 9.81 × 2.5
x
=
vₓ
×
t
25 = 10 × 2.5
y
=
h
+
v₀
×
sin(
θ
°)
×
t
−
1
2
×
9.81
×
t
2
14.645
=
2
+
20
×
sin(60°)
×
2.5
−
1
2
×
9.81
×
2.5
2
v
=
√
vₓ² +
vᵧ
2
12.325
=
√(
10
2
+
(−7.2045)
2
)
φ
=
tan⁻¹(
vᵧ
vₓ
)
−35.7708
=
tan⁻¹(−
7.2045
10
)
Assumptions
•
No air resistance, and g = 9.81 m/s² downward; x is across from the launch point, y is height above level ground.
•
Across nothing pushes, so vₓ = v₀ cos θ stays the same; vᵧ starts at v₀ sin θ and drops by 9.81 m/s every second.
•
The angle θ is measured from level, + above and − below; φ is the direction of the velocity, measured the same way.
Step-by-step
WE KNOW
v₀ = 20 m/s, θ = 60°, h = 2 m, t = 2.5 s
FIND
horizontal velocity (vₓ), vertical velocity at t (vᵧ), distance across at t (x), height at t (y), speed at t (v), direction at t (φ)
Step 1 · Find horizontal velocity (vₓ)
Use vₓ = v₀ × cos(θ°)
The across part of the launch velocity. Nothing pushes across, so it stays the same all flight.
vₓ = 20 × cos(60°)
vₓ = 20 × 0.5
vₓ = 10 m/s
Step 2 · Find vertical velocity at t (vᵧ)
Use vᵧ = v₀ × sin(θ°) − 9.81 × t
The up part of the launch velocity, less the 9.81 m/s gravity takes off it every second.
vᵧ = 20 × sin(60°) − 9.81 × 2.5
vᵧ = 20 × 0.866 − 9.81 × 2.5
vᵧ = 17.32 − 24.525
vᵧ = −7.2045 m/s
Step 3 · Find distance across at t (x)
Use x = vₓ × t
Across, the ball moves at a steady vₓ, so the distance is vₓ times the time.
x
=
10
×
2.5
x
=
25
m
Step 4 · Find height at t (y)
Use y = h + v₀ × sin(θ°) × t − ½ × 9.81 × t²
Start
at
h
,
rise
at
v₀
sin
θ
,
and
fall
1
2
gt²
below
that.
y
=
2
+
20
×
sin(60°)
×
2.5
−
1
2
×
9.81
×
2.5
2
y
=
2
+
20
×
0.86603
×
2.5
−
1
2
×
9.81
×
6.25
y
=
2
+
17.3206
×
2.5
−
4.905
×
6.25
y
=
2
+
43.3015
−
30.656
y
=
14.645
m
Step 5 · Find speed at t (v)
Use v = √(vₓ² + vᵧ²)
The two parts are at right angles, so the speed is their Pythagorean sum.
v
=
√(
10
2
+
(−7.2045)
2
)
v
=
√
100 + 51.905
v
=
√
151.905
v
=
12.325
m/s
Step 6 · Find direction at t (φ)
Use φ = tan⁻¹(vᵧ ÷ vₓ)
The angle whose tangent is vᵧ over vₓ: + means still rising, − means falling.
φ
=
tan⁻¹(−
7.2045
10
)
φ
=
−35.7708°
Check
Put the numbers back in. Both sides should match.
10 = 20 × cos(60°) ✓
−7.2045 = 20 × sin(60°) − 9.81 × 2.5 ✓
25 = 10 × 2.5 ✓
14.645
=
2
+
20
×
sin(60°)
×
2.5
−
1
2
×
9.81
×
2.5
2
✓
12.325
=
√(
10
2
+
(−7.2045)
2
)
✓
−35.7708
=
tan⁻¹(−
7.2045
10
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.