Asmall object begins a free-fall from a height of h = 86.5 m at t0 = 0 s. after ? = 2.80 s, a second small object is launched vertically up from the ground with the initial velocity v0 = 41.2 m/s. at what height from the ground will the two objects first meet?

This can be solve using the formulad = vt + 0.5at^2
where d is the distancev is the initial velocitya is the acceleration ( a = - 9.81 m/s2 due to gravity)t is the time
fisrt lets calculate the distance of the small object after 2.80 sd = 0(2.8) + 0.5(9.81)(2.8^2)d = - 38.4552 md = 86.5 - 38.4552 d = 48.0448 m
so at this point let as calculate the distance they are equal and solve the velocity of the small object at t = 2.8vf = v + atvf = 9.81(2.8) = 27.488d = d86.5 - 27.488t - 0.5at^2 = vt + 0.5at^20(t) + 0.5(9.81)t^2 = 41.2t + 0.5(-9.81)t^2solve for tt = 1.26 s
substitute td = = 41.2t + 0.5(-9.81)t^2d = 41.2(1.26) + 0.5(-9.81)(1.26)^2d = 44.11 m

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