Question

A stone is dropped freely , while another is thrown vertically downward with an initial velocity of 2m/s from the same point , simultaneously .what is the time required by them to have distance of separation 22m between them ???

Answer 1

Answer:

t=11 s

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

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Answer 2

Acceleration due to gravity be 10 m/s².

initial velocity of ball A = 0 m/s.
initial velocity of ball B = 2 m/s.

Using 2nd equation of uniformly accelerated motion.
we have, h = ut + 1/2gt².

distance covered by ball A,
h(A) = 1/2 × 10 × t = 5t

distance covered by ball B,
h(B) = 2 × t + 1/2 × 10 × t = 2t + 5t = 7t

According to question,
h(B) - h(A) = 22
7t - 5t = 22
2t = 22
t = 11 seconds.

Time required by them to have distance of separation 22m between them is 11 seconds.

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