Question

ABC is a triangle in vertical plane. Its two base angles $\angle BAC$ and $\angle BCA$ are $45^{\circ}$ and \[ta{{n}^{-1}}(1/3)\] respectively. A particle is projected from point A such that it passes through vertices $B$ and $C$. Find angle of projection in degrees:

(A) $60^{\circ}$
(B) $53^{\circ}$
(C) $\tan ^{-1}(5 / 4)$
(D) $\tan ^{-1}(5 / 3)$

Answer 1

Hint: We should know that velocity is defined as the rate change of displacement per unit time. Speed in a specific direction is also known as velocity. Velocity is equal to displacement divided by time. Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance which is a scalar quantity per time ratio. On the other hand, velocity is a vector quantity; it is direction-aware. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion in this case. Based on this we have to solve this question.

Complete step by step answer
We should know that a projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity. 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. A projectile is any object that has been thrown, shot, or launched, and ballistics is the study of projectile motion. Examples of projectiles range from a golf ball in flight, to a curve ball thrown by a baseball pitcher to a rocket fired into space.
The diagram for this question is given as:

Equation $y=x \tan \theta\left(1-\dfrac{x}{R}\right)$
at $B x=y$ $\tan \theta=\dfrac{R}{R-Y} \ldots.. (\mathrm{i})$
$\tan 45^{\circ}=\dfrac{Y}{X}$
$x=y \ldots.. (\mathrm{ii})$
$\left(\dfrac{1}{3}\right)=\dfrac{Y}{R-x} \ldots. (iii)$
Solving equation 2 and 3 $R=4y=4x$ it on $(i)$
$\tan \theta=\dfrac{R}{R-\dfrac{R}{4}}$
$\tan \theta=\dfrac{4}{3}$
$\theta=53^{\circ}$

So, option B is correct.

Note: We should know that if an object's speed or velocity is increasing at a constant rate then we say it has uniform acceleration. The rate of acceleration is constant. If a car speeds up then slows down then speeds up it doesn't have uniform acceleration. The instantaneous acceleration, or simply acceleration, is defined as the limit of the average acceleration when the interval of time considered approaches 0. It is also defined in a similar manner as the derivative of velocity with respect to time. If an object begins acceleration from rest or a standstill, its initial time is 0. If we get a negative value for acceleration, it means the object is slowing down. The acceleration of an object is its change in velocity over an increment of time. This can mean a change in the object's speed or direction. Average acceleration is the change of velocity over a period of time. Constant or uniform acceleration is when the velocity changes the same amount in every equal time period.