Question

A small object begins a free-fall from a height of ?=83.0 m at ?0=0 s . After τ=2.65 s , a second small object is launched vertically up from the ground with an initial velocity of ?0=40.4 m/s . At what height from the ground will the two objects first meet?

Answer 1

given that
Velocity of the falling object at t=2.65 sec
v = -g*t = -9.8*2.65 = -24.5 m/s
s = 0.5*g*t^2 = 0.5*9.8*2.65^2 = 34.4 m
so height from the ground of the falling object at t = 2.65 sec is 83 - 34.4 = 48.6 m
now relative to the object through up falling object will be approaching it with constant velocity {because acceleration of both object is same g downwards}
relative velocity = 40.4 - (-24.5) = 64.9 m/s
time = d/v = 48.6/64.9 = 0.748 sec
calculating height by the distance travelled by upward thrown object
h = ut + 0.5*a*t^2
h = 40.4*0.748 - 0.5*9.8*(0.748^2)
h = 27.479 m