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

The equation of trajectory of a profile is y=bxax2 (symbols have their usual meanings and a and b are positive constants). The angle of projection (from the horizontal) for the projectile is:
(A) tan1(b)
(B) tan1(a)
(C) cos1(b)
(D) cos1(a)

Answer
VerifiedVerified
460.2k+ views
like imagedislike image
Hint: The vertical and horizontal motions are independent of each other. Differentiate the equation of trajectory of a profile with respect to x and equate the result to zero. Find the value of x and compute it in the equation of trajectory of a profile.

Formula Used: The formulae used in the solution are given here.
 y=xtanθ where y is the vertical distance, x is the horizontal distance and the angle of projection is θ .

Complete step by step solution:
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motions along perpendicular axes are independent and thus can be analysed separately.
Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x and y , and the components of the velocity v are given by vx=vcosθ and vy=vsinθ , where v is the magnitude of the velocity and θ is its direction.
For horizontal motion, we have acceleration along the x-axis, ax=0 .
Using these values in the equations, of motion, we get,
 x=x0+vxt
 y=axbx2
It has been given that, the equation of trajectory of a profile is y=bxax2 where symbols have their usual meanings and a and b are positive constants.
We differentiate this equation of projectile with respect to x . We get,
 dydx=b2ax .
Thus, the angle of horizontal projection θ is given by,
 tanθ=dydx=a2bx
Now when we consider, dydx=0 ,
We have x=b2a .
Substituting x=b2a in the equation of trajectory of a profile, we have,
 y=b(b2a)a(b2a)2
 y=a22ba24b=a24b
Put, x=0 in the equation, tanθ=a2bx . Thus we get
 tanθ=a
 θ=tan1(a)
The angle of projection (from the horizontal) for the projectile is tan1(a) .
Hence the correct answer is Option B.

Note:
When we speak of the range of a projectile on level ground, we assume that range is very small compared with the circumference of the Earth. If, however, the range of the projectile is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits.

Latest Vedantu courses for you
Grade 10 | MAHARASHTRABOARD | SCHOOL | English
Vedantu 10 Maharashtra Pro Lite (2025-26)
calendar iconAcademic year 2025-26
language iconENGLISH
book iconUnlimited access till final school exam
tick
School Full course for MAHARASHTRABOARD students
PhysicsPhysics
BiologyBiology
ChemistryChemistry
MathsMaths
₹36,600 (9% Off)
₹33,300 per year
Select and buy