Kinematics Question 663
Question: Two identical particles are projected horizontally in opposite directions with a speed of $ 5\text{ m}{{\text{s}}^{-1}} $ each from the top of a tall tower as shown. Assuming $ \text{g = 10 m}{{\text{s}}^{-2}} $ , the distance between them at the moment when their velocity vectors become mutually perpendicular it’s
Options:
A) 2.5 m
B) 5 m
C) 10 m
D) 20 m
Show Answer
Answer:
Correct Answer: B
Solution:
[b] At a time t when velocity vector become mutually perpendicular $ vcos45{}^\circ =5 $ horizontal component $ v=\frac{5}{cos45{}^\circ }=5\sqrt{2}\text{ m/s } $ Vertically,
$ v\sin 45{}^\circ =gt $
$ \Rightarrow t=\frac{v\sin 45{}^\circ }{g}=\frac{5\sqrt{2}\times 1/\sqrt{2}}{10}=\frac{5}{10}=\frac{1}{2} $
So, $ OA=OB=v\cos 45{}^\circ \times t=5\times \frac{1}{2}=2.5 $
$ \Rightarrow AB=2.5\times 2=5\text{ m} $
