Kinematics Question 171
Question: 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 $ .
Options:
A) They meet at time $ t=\frac{u}{g} $ and at a height $ \frac{u^{2}}{2g}+\frac{gT^{2}}{8} $
B) They meet at time $ t=\frac{u}{g}+\frac{T}{2} $ and at a height $ \frac{u^{2}}{2g}+\frac{gT^{2}}{8} $
C) They meet at time $ t=\frac{u}{g}+\frac{T}{2} $ and at a height $ \frac{u^{2}}{2g}-\frac{gT^{2}}{8} $
D) They never meet
Show Answer
Answer:
Correct Answer: C
Solution:
[c] For first projectile, $ h _{1}=ut-\frac{1}{2}gt^{2} $ For second projectile,
$ h _{2}=u(t-T)-\frac{1}{2}g{{(t-T)}^{2}} $
When both meet i.e. $ h _{1} $ = $ h _{2} $ : $ ut-\frac{1}{2}gt^{2}=u(t-T)-\frac{1}{2}g{{(t-T)}^{2}} $
$ \Rightarrow uT+\frac{1}{2}gT^{2}=gtT\Rightarrow t=\frac{u}{g}+\frac{T}{2} $
And $ h _{1}=u( \frac{u}{g}+\frac{T}{2} )-\frac{1}{2}g{{( \frac{u}{2}+\frac{T}{2} )}^{2}}=\frac{u^{2}}{2g}-\frac{gT^{2}}{8} $
