Meet each other

Answer 1

Let h be the height where they meet each other after a time t sec So for falling body it has to cover a distance 156.8 - h m

==> S = ut + 1/2 g t^2

Here S = (156.8 - h) m, u = 0

So 156.8 - h = 1/2 g t^2

For rising body

s = h, u = 78.1 m/s and g = - g

So h = 78.1 t - 1/2 g t^2

Adding 156.8 = 78.1 t

So t = 2.008 s

Now you can find the value of h.

Answer 2

The distance traveled by the dropped object is s1 = 1/2 at^2 where a is gravity acceleration and t^2 is time squared.The distance traveled by the thrown object is s2 = vt where v is the average velocity(initial velocity/2)s1 + s2 = 156.8a = 9.81 m/s/sYou can now solve the two equations as1/2 at^2 +vt/2 -158 = 0t = 2.88 secondss1 = 114 m from tops2 = 42.4 m from bottom

Answer 3

After approx 2.01 seconds at a height of 137.05 metres.