Question
Question: A ball is dropped from a building of height 45 m. Simultaneously another identical ball B is thrown ...
A ball is dropped from a building of height 45 m. Simultaneously another identical ball B is thrown up with a speed 50ms−1. The relative speed of ball B w.r.t ball A at any instant of time is (Take g=10ms−2).
& A.\,0 \\\ & B.\,10\,m{{s}^{-1}} \\\ & C.\,25\,m{{s}^{-1}} \\\ & D.\,50\,m{{s}^{-1}} \\\ \end{aligned}$$Solution
The velocity of an object relative to the other object that might be stationary or moving with either same or a different velocity. The magnitude of the velocity of objects with respect to one another will be more than the individual velocities.
Formula used:
v=u−gt
v12=v1−v2
v21=v2−v1
Complete step by step answer:
The relative velocity can be defined as the rate of change of relative position of one object concerning another.
The relative velocity of object 1 concerning to object 2 is mathematically represented as follows. v12=v1−v2. Similarly, the relative velocity of object 2 concerning to object 1 is mathematically represented as follows. v21=v2−v1. Irrespective of the direction, the magnitude of these relative velocities will be equal, that is, ∣v12∣=∣v21∣.
From the given information we have the data as follows.
A ball is dropped from a building of height 45 m. Simultaneously another identical ball B is thrown up with a speed 50ms−1.
The formula that we will be using to find the relative velocity is,
v=u−gt
Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity and t is the time taken.
Consider the ball A.