187.
A stone is dropped freely, while another
thrown vertically downward with an initial
velocity of 2ms-l from the same point,
simultaneously. The time required by them
to have a distance of separation 22 m
between them is
(A) 11s
(B) 5.5 s
(C) 445
(D) 22s
Answers
Answer:
11 sec
Explanation:
We're asked to find the time t when the distance between the two stones, which we'll call Δy, is 22 m.
The equation for the position of stone 1 (the one dropped from rest) is
Δy1=−12gt2
(the initial y-velocity v0y is 0)
and that of stone 2 (the one launched downward) is
Δy2=v0yt−12gt2
Plugging in the known values, these equations become
Δy1=(−4.9ms2)t2
Δy2=(−2ms)t−(4.9ms2)t2
At any given time, the position of stone 2 will always be farther than that of stone 1, because it was launched with an initial velocity downward. Since we want to find when the distance between them is 22 m, we can say
Δy2=Δy1−22 m (minus because it's farther downward)
Substituting the above equations for Δy2 and Δy1, we have
(−2ms)t−(4.9ms2)t2=(−4.9ms2)t2−22 m
(−2ms)t=−22 m
t=11 s
Thus, the distance between the objects will be 22 meters after 11 seconds.
We can check to see if this answer is correct by finding their positions at time t=11 s:
Δy1=−12gt2=(−4.9ms2)(11s)2=−593 m
Δy2=v0yt−12gt2=(−2ms)(11s)−(4.9ms2)(11s)2
=−615 m
Distance between them=Δy1−Δy2=−593m−(−615m)
=22 m
Answer:
We're asked to find the time t when the distance between the two stones, which we'll call Δy, is 22 m.
The equation for the position of stone 1 (the one dropped from rest) is
Δy1=−12gt2
(the initial y-velocity v0y is 0)
and that of stone 2 (the one launched downward) is
Δy2=v0yt−12gt2
Plugging in the known values, these equations become
Δy1=(−4.9ms2)t2
Δy2=(−2ms)t−(4.9ms2)t2
At any given time, the position of stone 2 will always be farther than that of stone 1, because it was launched with an initial velocity downward. Since we want to find when the distance between them is 22 m, we can say
Δy2=Δy1−22 m (minus because it's farther downward)
Substituting the above equations for Δy2 and Δy1, we have
(−2ms)t−(4.9ms2)t2=(−4.9ms2)t2−22 m
(−2ms)t=−22 m
t=11 s
Thus, the distance between the objects will be 22 meters after 11 seconds.
We can check to see if this answer is correct by finding their positions at time t=11 s:
Δy1=−12gt2=(−4.9ms2)(11s)2=−593 m
Δy2=v0yt−12gt2=(−2ms)(11s)−(4.9ms2)(11s)2
=−615 m
Distance between them=Δy1−Δy2=−593m−(−615m)
=22 m
Explanation:
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