A body is thrown vertically up with the velocity of 100m/s and another

1.	A body is thrown vertically up with the velocity of 100m/s and another one is thrown 4sec after the first one how long after the first one is throw n will they meet (g =10m/s)
Answer» Given: A body is THROWN vertically up with the velocity of 100m/s and another one is thrown 4sec after the first one with same velocity. To find: After how much time does the bodies meet ? Calculation: 1st body reaches a POSITION after (t+4)sec. 2ND body reaches the same position after being thrown after 4 sec, so it's time is t sec. For 1st body: $\therefore \: h = u(t + 4) - \dfrac{1}{2} g {(t + 4)}^{2} . \: . \: . \: .(1)$ For 2nd body: $\therefore \: h = ut - \dfrac{1}{2} g {t}^{2} \: . \: . \: . \: .(2)$ EQUATING the two EQUATIONS: $\implies \: u(t + 4) - \dfrac{1}{2} g {(t + 4)}^{2} = ut - \dfrac{1}{2} g {t}^{2}$ $\implies \: 4u - \dfrac{1}{2} g {(t + 4)}^{2} = - \dfrac{1}{2} g {t}^{2}$ $\implies \: \dfrac{8u}{g} = {(t + 4)}^{2} - {t}^{2}$ $\implies \: \dfrac{8u}{g} = 8t + 16$ $\implies \: 8t + 16 = \dfrac{8 \times 100}{10}$ $\implies \: 8t + 16 = 80$ $\implies \: 8t = 64$ $\implies \: t = 8 \: sec$ So, bodies meet after (8 + 4) = 12 sec.