Question

A small object begins a free-fall from a height of H = 86.5 m at t0 = 0s. After tau = 2.65s, a second small object is launced vertically up from the ground with the initial velocity v0 = 41.2 m/s. At what height from the gound will the two objects first meet?

Answer 1

Let the objects meet after time 't'

time of travel for first object = 2.65 + t

time of travel for second object = t

distance travelled by the object dropped is given as ::

d₁ = v_1i t + (0.5) a t²

d₁ = (0) (t + 2.65) + (0.5) (9.8) (2.65 + t)²

d₁ = 4.9 (2.65 + t)²             eq-1

distance travelled by the object launched from ground is given as ::

d₂ = v_2i t + (0.5) g t²

d₂ = 41.2 t - (0.5) (9.8) t²           Eq-2

d₁ + d₂ = 86.5

4.9 (2.65 + t)² + 41.2 t - (0.5) (9.8) t² = 86.5

t = 0.76 sec

height from ground is given as ::

d₂ = 41.2 t - (0.5) (9.8) t²  

d₂ = 41.2 (0.76) - (0.5) (9.8) (0.76)²  

d₂ = 28.48 m