Acceleration, velocity, displacement of projectile at different points of its trajectory

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Acceleration, velocity, displacement of projectile at different points of its parabolic trajectory are what we are going to study here. Both horizontal and vertical components of these 3 motion quantities will be presented in tabular format.

finding the acceleration, velocity and displacement of a projectile at different points on the parabolic trajectory of projectile
figure 1

Consider a ball that is projected at an angle θ to the horizontal with velocity v, as shown in Figure 1. R is the horizontal range. We can split the motion into three parts, beginning, middle, and end, and analyze the vectors representing displacement, velocity, and time at each stage.

Note that since the path is symmetrical, the motion on the way down is the same as the way up.

We can see that the vertical motion is constant acceleration and the horizontal motion is constant velocity.

Horizontal components of displacement, velocity, and acceleration of a projectile

figure 2

Consider a ball that is projected at an angle θ to the horizontal with velocity v, as shown in Figure 2.
R is the horizontal range.
At A (point of projection) the horizontal displacement is zero, the horizontal velocity is v cos θ and horizontal acceleration is 0.
At B (highest point of trajectory) the horizontal displacement is R/2, the horizontal velocity is v cos θ and horizontal acceleration is 0.
And at C (last point of trajectory) the horizontal displacement is R, the horizontal velocity is v cos θ and horizontal acceleration is 0.
These are presented in table 1 below. Refer to figure 2 while going through the table data.
We can easily say that the horizontal motion is constant velocity.

At A (time = 0)At B (time = t /2)At C (time = t)
Displacement = zeroDisplacement = R/2Displacement = R
Velocity = v cos θVelocity = v cos θVelocity = v cos θ
Acceleration = 0Acceleration = 0Acceleration = 0
Table 1: Horizontal components of displacement, velocity, and acceleration of a projectile

Vertical components of displacement, velocity, and acceleration of a projectile

figure 3

Consider a ball that is projected at an angle θ to the horizontal with velocity v, as shown in Figure 3.
R is the horizontal range.
At A (point of projection) the vertical displacement is 0, the vertical velocity is v sin θ and vertical acceleration is -g.
At B (highest point of trajectory) the vertical displacement is h, the vertical velocity is 0 and vertical acceleration is -g.
And at C (last point of trajectory) the vertical displacement is 0, the vertical velocity is -v sin θ and vertical acceleration is -g.
These are presented in table 2 below. Refer to figure 3 while going through the table data.
We can see easily that the vertical motion is constant acceleration.

At A (time = 0)At B (time = t /2)At C (time = t)
Displacement = zeroDisplacement = hDisplacement = 0
Velocity = v sin θVelocity = 0Velocity = -v sin θ
Acceleration = -gAcceleration = -gAcceleration = -g
Table 2: Vertical components of displacement, velocity, and acceleration of a projectile

Also read: derivation of the projectile equation & important FAQs

At the highest point of the trajectory, what is the acceleration of a projectile?

The acceleration of the projectile at the peak is -g. g denotes the acceleration due to gravity (9.81 m/s^2).

The minus sign indicates that the direction of acceleration of the projectile at the highest point of trajectory is opposite to the positive Y-axis, i.e. its direction is towards the earth. g denotes the acceleration due to gravity.

[Note: At the highest point, the horizontal component of the acceleration is zero, and the vertical component of acceleration is -g. ]

Vertical motion is constant acceleration for a projectile – how?

The vertical component of acceleration remains constant at -g during the entire trajectory of the projectile. Hence, the vertical motion of a projectile is constant acceleration.

Also read: Numerical problems on projectile motion

Horizontal motion is constant velocity for a projectile – how?

The horizontal component of the acceleration of the projectile is 0 and hence the projectile maintains a constant velocity in its horizontal motion.

What is the velocity of a projectile at the peak of its path?

figure 4

At the top of its flight, when the object stops rising and is about to fall, the vertical speed of an object is zero. The horizontal speed at that point is expressed as v cos θ.

(Considering that it is projected at an angle θ to the horizontal with velocity v.
Refer to figure 4)

Related:
1> derivation of the projectile equation & important FAQs
2> Numerical problems on projectile motion

Acceleration, velocity, displacement of projectile at different points of its trajectory
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