Kinematics Question 416
A body a is thrown vertically upward with the initial velocity $ v _{1} $ . Another body b is dropped from a height h. Find how the distance x between the bodies depends on the time t if the bodies begin to move simultaneously.
Options:
A) $ x=h-v _{1}t $
B) $ x=( h-v _{1} )t $
C) $ x=h-\frac{v _{1}}{t} $
D) $ x=\frac{h}{t}-v _{1} $
Show Answer
Answer:
Correct Answer: A
Solution:
The distance travelled by the body is A,
given by $ v _{1}t=-\frac{gt}{2} $ and that travelled by the body b is $ h _{2}=\frac{gt^{2}}{2} $ s
The distance between the bodies $ =x=h-(h _{1}+h _{2}). $ Since $ h _{1}+h _{2}={v _{1}}t $,
the relation sought is $ x=h-v _{1}t $
BETA
