An object thrown at an angle moves at constant velocity horizontally and, because of gravity, with constant acceleration vertically, both at once.
The two motions do not affect each other and share only the time; for the same speed the range is longest at a launch angle of 45°, and two objects thrown at θ and 90° − θ land at the same spot.
At the highest point the vertical velocity becomes 0, but the horizontal velocity is still there, so the speed does not become 0.
In the picture, changing the launch angle and the initial speed changes the path along with the range, the maximum height and the speed at the highest point.
Horizontal and Vertical Motion Happen Separately
🎯 Independence of Motion
①An object thrown at an angle moves horizontally and vertically at the same time
②Horizontally there is no force, so the velocity is constant (uniform motion)
③Vertically only gravity acts, so the acceleration is a constant g (uniformly accelerated motion)
④The two motions do not affect each other and share only the time t
The Path for Each Angle and Initial Speed
45°
20 m/s
seegongsik.com
The parabolic path of an object thrown from the ground (g = 10 m/s²). The green dot is the highest point, and the green arrow is the horizontal velocity that remains there
📐 Why Does 45° Go Farthest?
①The range R = v₀² sin2θ/g is longest at θ = 45°, where sin2θ reaches its largest value, 1
②θ and 90° − θ give the same sin2θ and so the same range (for example 30° and 60°)
③A larger launch angle sends the object higher and keeps it in the air longer, but the horizontal velocity shrinks, so past 45° the range gets shorter
Horizontal and Vertical Motion as Equations
Horizontal: Uniform Motion
x = v0 cosθ · t, vx = v0 cosθ
No force acts horizontally, so the horizontal velocity does not change
Vertical: Uniformly Accelerated Motion
y = v0 sinθ · t − 12gt2, vy = v0 sinθ − gt
On the way up gravity reduces vy, and on the way down it increases it
Range, Maximum Height and Time of Flight
Horizontal Range
R = v02 sin2θg
For the same initial speed, it is longest at θ = 45°
Maximum height
H = (v0 sinθ)22g
The height at which vy = 0, the highest point
Time of flight
T = 2 v0 sinθg
The time up equals the time down, so the time of flight is twice the time up
Angle comparison
Comparing launch angles at v₀ = 20 m/s (g = 10 m/s²)
Angle
Range
Max height
Flight time
30°
34.6 m
5 m
2 s
45°
40 m
10 m
2.8 s
60°
34.6 m
15 m
3.5 s
From Concept to Problem
Example 1: Finding the Range
An object is thrown with an initial speed of 20 m/s at a launch angle of 45°. Find its horizontal range. (g = 10 m/s²)
1
Put the values into the range formula.
R = v02 sin2θg
2
Use sin(2 × 45°) = sin 90° = 1.
R = 202 × 110 = 40
▸ 40 m
At 45°, sin2θ reaches its largest value, 1, so this angle gives the longest range for the same speed.
Example 2: Speed at the Highest Point
An object is thrown with an initial speed of 20 m/s at a launch angle of 60°. Find its speed and direction of motion at the highest point.
1
At the highest point the vertical velocity vy = 0, so only the horizontal component of the speed remains.
2
Find the horizontal velocity.
vx = v0 cosθ = 20 × cos60° = 20 × 0.5 = 10
▸ Speed 10 m/s, moving horizontally
Even at the highest point the horizontal velocity is the same as at launch, so the speed is not 0. The green arrow in the picture is exactly this velocity.
Exam Points
CSAT-style problem (Physics II), adapted
With the same initial speed, object A is thrown at 30° and object B at 60°. Choose the correct statement (ignore air resistance).
①A has a larger range than B.
②A and B have the same range.
③A rises higher than B.
④A has a longer time of flight than B.
⑤At the apex the speed of both objects is zero.
▸ ② A and B have the same range.
1
The range is R ∝ sin2θ, and sin(2 × 30°) = sin 60° and sin(2 × 60°) = sin 120° = sin 60° are equal, so ② is correct.
2
The maximum height is H ∝ sin²θ, so B, thrown at 60°, goes higher and stays in the air longer. So ③ and ④ are wrong.
3
At the highest point the horizontal velocity v₀cosθ remains, so the speed is not 0, and ⑤ is wrong too.
🎯 Exam Points
①Horizontal motion is uniform (v₀cosθ) and vertical motion is uniformly accelerated by gravity; the two are independent
②The range R = v₀² sin2θ/g is longest at θ = 45° and the same for θ and 90° − θ
③The maximum height is H = (v₀sinθ)²/2g, and v_y = 0 at the highest point
④Time of flight T = 2v₀sinθ/g
⑤The speed at the highest point is not 0 but the horizontal component v₀cosθ