A ball is dropped from the top of a building and simultaneously, another ball is projected vertically upward from the ground with a velocity of 25 m/s. If the initial distance between these balls is 100 m. Calculate where the two balls will meet above the ground.
(Take g = 10 m/s2 )
Initial velocity of ball(b1) = 0 m/s
Initial velocity of ball(b2) = 25 m/s
Distance covered by ball(b2) in time t = x
Total height of the cliff(H) = 100 m
●Distance covered by ball(b1) in time t = 100 − x
Plug in values in second equation of motion
For Ball (b1) :
100 − x = 0 + 12gt2
100 − x = 12gt2 ....(1)
For Ball(b2) :
x = u2t - 12gt2 …(2)
On adding (1) & (2), we get the final equation :100 =12gt2 +u2t - 12gt2
100 = u2t
t = 4 s
Substituting t = 4 in equation 1
x = (25)(4) - 12(10∗42)
x = 20 m