Question

A particle projected from a horizontal plane ( $ x - y $ plane) such that its velocity vector at time $ t $ is given by $ \vec V = a\hat i + \left( {b - ct} \right)\hat j $ . Its range on the horizontal plane is given by
(A) $ \dfrac{{ba}}{c} $
(B) $ \dfrac{{2ba}}{c} $
(C) $ \dfrac{{3ba}}{c} $
(D) None

Answer 1

Hint :From the given equation of velocity, the velocity and acceleration components in the horizontal and vertical direction can be found out. We need to find the range on the horizontal plane, which is the distance traveled in the horizontal plane which is the product of the horizontal component of velocity and the time taken to travel.

Complete Step By Step Answer:
Let us note the given data and derive the given values,
 $ \vec V = a\hat i + \left( {b - ct} \right)\hat j $
We know that the $ \hat i $ shows the $ x $ - component and $ \hat j $ shows the $ y $ - component of the velocity
Thus, from the given data, we can derive
 $ {v_x} = (a)m{s^{ - 1}} $
 $ {v_y} = (b - ct)m{s^{ - 1}} $
The second term of the $ y $ component of velocity contains time $ t $ , which means that $ - c $ shows the acceleration in the vertical direction.
 $ \therefore {a_y} = ( - c)m{s^{ - 2}} $
The particle moves in a projectile motion, which means it started its motion from the ground i.e. $ y = 0 $ and came back to the ground after traveling the distance equal to range in horizontal direction i.e. $ y = 0 $ at the end.
Now, from the kinematic equation of uniform acceleration,
 $ s = {u_i}t + \dfrac{1}{2}a{t^2} $
Considering the equation for motion in $ y $ - direction
 $ y = {v_y}t + \dfrac{1}{2}{a_y}{t^2} $
Substituting the values as per the given data for the condition when particle comes back to the ground at the end i.e. $ y = 0 $
 $ \therefore 0 = bt - \dfrac{1}{2}c{t^2} $
 $ \therefore bt = \dfrac{1}{2}c{t^2} $
Canceling the common terms,
 $ \therefore b = \dfrac{1}{2}ct $
 $ \therefore \dfrac{{2b}}{c} = t $
The range on the horizontal plane is the product of velocity in the horizontal direction and the time taken to travel the distance.
 $ \therefore $ Range = $ {v_x} \times t $
Substituting the derived values,
 $ \therefore $ Range = $ a \times \dfrac{{2b}}{c} $
 $ \therefore $ Range in the horizontal direction = $ \dfrac{{2ba}}{c} $
The correct answer is Option $ (B) $ .

Note :
In the given question, the horizontal component of velocity only has a velocity value, which means that acceleration is zero in the horizontal direction. As we know that acceleration is velocity per unit time, velocity can be written as the product of acceleration and time. In the second term of the vertical component of velocity, we have a similar product. By this, we can obtain the value of acceleration.