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

Well the magnitude of its acceleration related to that car starts with the oil change like when you have to change the oil in your car you have to make sure that your precious car is starting really propley and very well