Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

A large number of bullets are fired in all directions with the same speed u. What is the maximum area on the ground on which these bullets will spread?
A.)πu2g
B.)πu4g2
C.)π2u4g2
D.)π2u2g2


Answer
VerifiedVerified
508.2k+ views
like imagedislike image
Hint: Try and interpret the problem as a case of projectile motion. The projectiles together form a circular distribution on the ground. The radius of this circle will be the maximum range of the projectile, i.e., the launch angle of the projectile θ=45.

Formula used:
Area of circle = πr2, where r is the radius of the circle.
Range of a projectile R=u2sin2θg where u is the launch velocity, θ is the launch angle, and g is the acceleration due to gravity.

Complete answer:
Let us deconstruct the question to understand it in terms that are familiar to us.
The question basically deals with a case of projectile motion. Projectile motion is a form of motion where an object moves in a parabolic path. It occurs when there is one force that launches the object after which the only influencing force is gravity. The path that the object follows under this influence of gravity is called projectile trajectory. In most cases for our convenience we neglect air resistance.
A bullet fired from a gun is an example of projectile motion.
The range of a projectile can be understood as the horizontal distance between the launch point and the point where the projectile hits the ground.

Now, let us try and understand what the question entails.
We have a gun that fires bullets in all directions. This means that the distribution of all bullets landing on the ground would be in the form of a circle, since they’re all fired with the same launch velocity u, and the radius of this circle would be equal to the range of the projectile.

Therefore, the area covered by the bullets on the ground can be given as A=πR2
The range of a projectile is given as R=u2sin2θg.
Now, the question asks for us to find the “maximum” area that can be obtained from the projectile distribution. Therefore, we need the maximum range that the projectiles can achieve.
From the equation for range R we see that the maximum value of R that we can get is

Rmax=u2g where the value of sin2θ=1θ=45
Substituting Rmax in area, we get:
Amax=π×Rmax2πu4g2

So, the correct answer is “Option B”.

Note:
Here, we have assumed that the bullets are launched from the flat ground. However, if you should encounter any projectile that is launched from a certain height h, then the range of the projectile takes the form:
R=u22g(1+1+2ghu2sinθ)sin2θ
Latest Vedantu courses for you
Grade 10 | CBSE | SCHOOL | English
Vedantu 10 CBSE Pro Course - (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for CBSE students
PhysicsPhysics
Social scienceSocial science
ChemistryChemistry
MathsMaths
BiologyBiology
EnglishEnglish
₹41,000 (9% Off)
₹37,300 per year
Select and buy