A projectile is fired vertically upwards with an initial velocity u. After an interval of T seconds a second projectile is fired vertically upwards, also with initial velocity u.

(1) They meet at time $t=\frac{u}{g}$ and at a height $\frac{{u}^{2}}{2g}+\frac{g{T}^{2}}{8}$

(2) They meet at time $t=\frac{u}{g}+\frac{T}{2}$ and at a height $\frac{{u}^{2}}{2g}+\frac{g{T}^{2}}{8}$

(3) They meet at time $t=\frac{u}{g}+\frac{T}{2}$ and at a height $\frac{{u}^{2}}{2g}-\frac{g{T}^{2}}{8}$

(4) They never meet

Concept Videos :-

#26-Equations-for-Uniform-Acceleration
#27-Equations for Uniform Acceleration 28Calculus Method29
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Concept Questions :-

Uniformly accelerated motion
Relative motion in 1-D

(3) For first projectile, ${h}_{1}=ut-\frac{1}{2}g{t}^{2}$

For second projectile, ${h}_{2}=u\left(t-T\right)-\frac{1}{2}g{\left(t-T\right)}^{2}$

When both meet i.e. ${h}_{1}={h}_{2}$

$ut-\frac{1}{2}g{t}^{2}=u\left(t-T\right)-\frac{1}{2}g{\left(t-T\right)}^{2}$

$uT+\frac{1}{2}g{T}^{2}=gtT$

$t=\frac{u}{g}+\frac{T}{2}$

and ${h}_{1}=u\left(\frac{u}{g}+\frac{T}{2}\right)-\frac{1}{2}g{\left(\frac{u}{2}+\frac{T}{2}\right)}^{2}$

$=\frac{{u}^{2}}{2g}-\frac{g{T}^{2}}{8}$

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