Horizontal Projectile Motion

Horizontal Projectile Motion

Let's start with some important equations, which are initial velocity after t sec, displacement acceleration and path of projectile and more.

Important Equations

• Initial Velocity (u):

Horizontal component =ux=u
Vertical component =uy=0

Velocity 'v' after time 't' sec-

Horizontal component =vx=u
Vertical component =vy=g.t
and,
v=vx2+vy2 i.e; v=u2+(gt)2tan⁡β=gtu

Where, β= angle that velocity makes with horizontal

• Displacement (S):

Horizontal component =Sx=u.t
Vertical component =Sy=12g⋅t2
and, S=Sx2+Sy2
Acceleration =a
Horizontal component =0
Vertical component =g
So, a=g

So, a = g

• Equation of path of a projectile

y=g2u2⋅x2
g→ Acceleration due to gravity
u→ initial velocity

Till now we have studied the important equation now coming to the important terms.

Important Terms

• Time of flight

t=2hg

where t is the time of flight and h= Height from which the projectile is projected.

• Range of Projectile

R=u⋅2hg

Where R is the Range of projectile and u is the horizontal velocity of projection from height h

• Velocity at which projectile hit the ground

v=u2+2gh

Where v is the velocity at which the projectile hit the ground.

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Solved Example Based on Horizontal Projectile Motion

Example 1: A particle is projected with a speed of 4m/s along a horizontal direction from a height. The equation of its path is:

1) y=513x2
2) y=1316x2
3) y=516x2
4) y=3x2

Solution:

Given:

x=4t−(1)y=12gt2=5t2−(2)

From 1 and 2:
y=516x2

Hence, the answer is option (3).

Example 2: A particle is projected in the horizontal direction from a height. The initial speed is 4m/s. Then the angle made by its velocity with horizontal direction after 1 second is :

1) tan−1⁡(25)
2) tan−1⁡(52)
3) tan−1⁡(12)
4) tan−1⁡(13)

Solution:

For Projectile Projected Horizontally -

tan⁡β=gtu
where
β= the angle that velocity makes with horizontal
So Angle with horizontal is
θ=tan−1⁡(gtu)⇒θ=tan−1⁡(104)=tan−1⁡(52)

Hence, the answer is option (2).

Example 3: A helicopter flying horizontally with a speed v at an altitude h has to drop a food packet for a man on the ground. What is the distance of the helicopter from the man when the food packet is dropped?

1) 2v2hg+h2
2) 2ghv2+h2
3) 2ghv2+h2
4) 2ghv2+1h2

Solution:

x=vt=v×2hg
r→ distance of man and helicopter
r=x2+h2r=v2×2hg+h2
Hence, the answer is option (1).

Example 4: A stone is thrown horizontally with a velocity of 10 m/s. Find the radius of curvature of its trajectory at the end of 2 seconds after motion began (g=10 m/s2)

1) 10010 m
2) 505 m
3) 5 m
4) 235m

Solution:

y=−gx22u2dydx=−gx2u2=d2ydx2=−gu2 After t=2secx=uxt=20 mdydx=−20g(200)2=−2d2ydx2=−10100=−110R=[[1+(dydx)2]3/2]d2ydx2]=[(1+(−2)2)3/2−110]=53/2110=505

Hence, the answer is the option (2).

Example 5: A child stands on the edge of the cliff 10 m above the ground and throws a stone horizontally with an initial speed of 5 ms−1. Neglecting the air resistance, the speed with which the stone hits the ground will be ms−1 (given, g=10 ms−2 ).

1)15

2)20

3)30

4)25

Solution:

Along vertical direction

vy2=uy2+2aygy=(0)2+2×10×10vy2=200

sy=10 mvy2=200∴v=vx2+vy2=25+200=225=15 m/s

Hence, the answer is option (1).

Summary

In conclusion, the basic underlying idea that covers the core of this concept is that of the horizontal projectile motion in physics. Utilizing initial horizontal velocity, this particular type of motion includes the constant horizontal speed of the object and the vertical acceleration known to change the speed, resulting in a parabolic trajectory. Understanding the principles of horizontal projectile motion, one will realize that there are different ways to use the concept in real life to achieve precision and safety in certain activities, for example, in designing rides in amusement parks or in ensuring that projectiles hit their targets with precision. It gives knowledge that will come in handy in various life situations to solve practical problems and create innovations.