Since Q is projected after 1 second, P will reach the maximum height and meet Q when it is returning back.
Let maximum height reached by P is H.
H = (u×u) /(2×g), where u is initial projection velocity 30 m/s
H = (30×30) / (2*10) = 45 m.
after reaching the maximumheight let P travels back for t seconds and meet Q.
Let the distance travelled by P in that t seconds is h and it is given by
h = (1/2)g×t2 = (1/2)×10×t2 = 5×t2 .........................(1)
Then distance travelled by Q when it meets P is 45-h and the time taken is (2+t)s. This distance is given by
45-h = 30×(2+t) - (1/2)×10×(2+t)2
after substituting for h from eqn.(1), we will get
45 - 5×t2 = 30×(2+t)-5×(2+t)2
solving the above eqn for t, we get t = 0.5 s.
Hence P travelled 3.5 s before meetin Q