A ball is projected horizontally from the top of a building. One second later, another ball is projected horizontally from the same point with the same velocity. At what point in the motion will the balls be closest to each other

t = 0 at the start of the projection

Step-by-step explanation:

To solve this we need to find the distance between the 2 positions at any given time, then solve for the least distance

Let t be the time of the 2nd ball, so t + 1 is the time of the first ball

Let g be the gravitational acceleration, v be the horizontal velocity

the y coordinates of the first and 2nd balls

The x coordinates of the 1st and 2nd balls:

The distance between the 2 balls is

As both v and g are constant and cannot be changed, d is minimum when (2t + 1) is minimum, which happens only when t is minimum = 0

the 4th choice is the correct one

step-by-step explanation:

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)