# Horizontal Speed Vs Horizontal Velocity: Comparative Analysis

The article discusses comparative analysis and insight about Horizontal Speed Vs Horizontal Velocity.

In the previous article, we learned that the projectile has only downward acceleration or vertical acceleration at all trajectory points due to gravity force, no matter how it is launched. On the other hand, the horizontal speed or horizontal velocity remains unchanging throughout the trajectory as no force accelerates the projectile horizontally.

Before the projectile attains its maximum height when launched at a certain angle, its vertical speed drops since the vertical acceleration is downward or opposite. The vertical speed becomes zero at maximum height, and the projectile carries with horizontal speed. But after maximum height, its vertical speed rises since it is in the direction of vertical acceleration.

The launch angle of the projectile decides its maximum height and its maximum horizontal distance, which depend on its time in the air. Therefore, the launch angle, mainly close to 45°, delivers the maximum horizontal range to the projectile if its initial speed is the exact as the horizontal speed. The launch angle has a good balance of the projectile’s initial velocity, optimizing its horizontal velocity and time in the air.

## How to differentiate Horizontal Speed and Horizontal Velocity?

Let’s discuss some projectile examples that differentiate between the projectile’s horizontal speed and horizontal velocity.

Suppose the airplane moves at a speed of 100m/s at an altitude of 1000m, dropping the box with a 10km/hr velocity towards the ground. In such a case, we note that the airplane moves in a straight direction with a constant horizontal speed of 100m/s. In distinction, the box falls in a curved trajectory when dropped from the airplane.

The falling box displays projectile motion, and it falls to the right with horizontal velocity and downward with vertical velocity due to gravity force. We also observe that the falling box is initially horizontal x-direction and not vertical y-direction. So, vy = 0. The horizontal velocity maintains its initial value vx = 100km/hr throughout the drop.

The airplane travels with constant horizontal speed above the falling box. The pilot always notices the falling box underneath the airplane, which illustrates that the falling box has no horizontal acceleration ax = 0. But the air resistance opposes the box’s velocity, which is why the falling box is not underneath the airplane as the pilot sees. When the box finally reaches the ground with acceleration ay= – 9.8m/s2, its y-component of displacement is y = -1000m.

Since horizontal velocity constant, we only obtain kinematics equations in the vertical velocity.

y = vyt + 1/2 ay t2

Since vy = 0, y = ay t2

The time for the falling box in air is given by

t = √(2y/ay)

Substituting all values,

t = √2(-1000)/-9.8

t = 14.28s

The fall time for the box is 14.28sec.

Suppose a boy throws the ball downward from the height as case A and the same boy throws the ball in a trajectory path towards the ground from the same height as case B. The speed is just the magnitude of the velocity. The velocity of the falling ball has an x-component i.e, horizontal velocity in case B, whereas it does not have in case A. Therefore, the ball falls with zero horizontal speed in case A whereas it falls with maximum horizontal velocity in case B.

During the javelin throw, the athlete must run some distance while carrying the javelin. When an athlete runs, the javelin also achieves the same horizontal speed as the athlete. In such a case, both the javelin and athlete’s vertical speed or vertical velocity stays zero as the athlete runs horizontally.

But when the athlete throws the javelin into the air, the gravity force acts on it, which yields the projectile motion. That’s why the javelin travels in parabolic trajectory with horizontal velocity initially and then the vertical velocity. The maximum height reached by the javelin depends on the launch angle from the horizontal axis and the vertical velocity of the javelin.

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