Projectile Motion for Horizontal Displacement

Suppose an object is thrown horizontally at a certain angle θ with a certain velocity. In that case, the path followed by the object is called trajectory, the thrown object is called projectile and the motion followed by it is called projectile motion. Once launched, the projectile moves by its inertia under the influence of gravitational force (the only force acting on the projectile is gravitational force).

How to find horizontal displacement in projectile motion?

If an object (called a projectile) is thrown with a certain constant velocity horizontally to the earth’s surface, and it moves along a parabolic path under the influence of gravity alone in the X-axis then it is known as horizontal projectile motion. Then the horizontal displacement due to the motion of the projectile is represented by the formula,

∆x = v_c × t

where

• ∆x = Horizontal displacement of projectile,
• v_c = Constant velocity of projectile, and
• T = Time of flight

Let’s imagine if an object is projected with an initial velocity v at an angle θ, and the initial height is h. Then, its components of velocity are given as, 

The horizontal velocity component: v_x = v.cos(θ) and 

The vertical velocity component: v_y = v.sin(θ).

In the case of the horizontal motion of the projectile, the vertical component of velocity is 0, therefore vsinθ is also 0.

Sample Questions

Question 1: Mohan travels 6 km to the North but then back-tracks to the South for 4 km to reach his home. Then total displacement of Mohan?

Answer:

The displacement of Mohan is given by the formula:

Δx = (x_f – x_i)

Here the initial position of Mohan is x_i

The final position is x_f which is the distance traveled towards the north minus the distance traveled towards the south.

Δx = (6 km North  – 4 km South) – 0

= 2 km North

Question 2: Find the value of horizontal displacement of a fired projectile at a constant velocity of 12 m/s and time of flight is 20 s.

Answer:

Given,

The velocity of particle = 12 m/s

Time of flight = 20 sec

Then Horizontal displacement of projectile is,

∆x = v_c x t

= 12 x 20 

= 240 m

Question 3: Find the value of horizontal displacement of a fired projectile at a constant velocity of 40 m/s and time of flight is 12 s.

Answer: 

Given,

Velocity of particle = 40 m/s

Time of flight = 12 sec

Then Horizontal displacement of projectile is,

∆x = v_c x t

= 40 x 12 

= 480 m

Question 4: A stone is thrown at a constant velocity of 10 m/s horizontally to the surface of the earth and the stone follows the parabolic path If the horizontal displacement attained by the stone is 100 m then find its time of flight.

Answer:

Given,

Velocity of stone = 10 m/s

Horizontal displacement of projectile = 100m 

Since,

∆x = v_c x t

Therefore,

Time of flight, t = ∆x / v_c

= 100 /10 

= 10 s 

Question 5: A boy kicked a football with a certain constant velocity v. If the ball follows a trajectory path with a horizontal displacement of 330 m and its time of flight during the projection is 20 s. Then what is the velocity of the ball?

Answer:

Given,

The velocity of football = v m/s

Time of flight = 20 sec

Horizontal displacement of football = 330 m

∆x = v_c x t

v_c = ∆x / t  

= 330 / 20 

= 16.5m/s