Question:

From a building two balls $A$ and $B$ are thrown such that $A$ is thrown upwards and $B$ downwards with the same speed (both vertically). If $v_A$ and $v_B$ are their respective velocities on reaching the ground then,

Updated On: Jun 14, 2022
  • $v_B > v_A$
  • $v_B = v_A$
  • $v_A > v_B$
  • their velocities depend on their masses
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Let the ball $A$ is thrown vertically upwards with speed $u$ and ball $B$ is thrown vertically downwards with the same speed $u$.
After reaching the highest point, $A$ comes back to its point of projection with the same speed $u$ in the downward direction. If $h$ be height of the building, then velocity of $A$ on reaching the ground is
$v^{2}_{A}=u^{2}+2gh$ or
$v_{A}=\sqrt{u^{2}+2gh}\,...\left(i\right)$
and that of $B$ on reaching the ground is
$v^{2}_{B}=u^{2}+2gh$ or
$v_{B}=\sqrt{u^{2}+2gh}\,...\left(ii\right)$
From eqns. $(i)$ and $(ii)$, we get $v_A = v_B$
Was this answer helpful?
0
0

Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration