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30 Jul 2019, 23:47
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If a ball is thrown from ground with velocity u m/s upwards in gravitation field with g \(m/s^2\) acceleration working downwards. The ball is rebounded and for each rebound the velocity reduces to f times of previous velocity (0<f<1). What is the total distance travelled by the ball since the time it is thrown upwards for the first time.
\(A. \frac{u^2}{g(1-f^2)}\)
\(B. \frac{u^2}{g(1-f)}\)
\(C. \frac{u}{g(1-f^2)}\)
\(D. \frac{u^2}{gf^2}\)
\(E. \frac{u}{gf^2}\)
\(A. \frac{u^2}{g(1-f^2)}\)
\(B. \frac{u^2}{g(1-f)}\)
\(C. \frac{u}{g(1-f^2)}\)
\(D. \frac{u^2}{gf^2}\)
\(E. \frac{u}{gf^2}\)
01 Aug 2019, 00:27
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from the third equation of motion
we know that--- \(v^2\) = \(u^2\)+2as ( where v= final velocity u= initial a= acceleration acting on body t= time s= distance covered)
when the body is thrown upward with initial velocity = u
its final velocity becomes zero when it reaches its highest point so v=0
acceleration acting on the body in the downward direction = -g m/s^2
\(0^2\)= \(u^2\)-2gs
s = \(\frac{u^2}{2g}\)
now it comes down so total distance = \(\frac{2u^2}{2g}\) (distance going up+distance coming down)
again it rebounds and goes upward with velocity = uf
total distance covered going up and coming down = \(\frac{2u^2f^2}{2g}\)
again it rebounds and goes upward with velocity = u\(f^2\) (we can find distance covered each time by above process)
this process continues till infinity
total distance covered during this process = \(\frac{2u^2}{2g}\)+\(\frac{2u^2f^2}{2g}\)+\(\frac{2u^2f^4}{2g}\)+............till infinity
total distance = \(\frac{u^2}{g}\)(1+\(f^2\)+\(f^4\)+\(f^6\)+............∞) (we see this forms geometric progression)
total distance = \(\frac{u^2}{g}\)(\(\frac{1}{1-f^2}\))....(sum of infinite GP with first term a and common ratio r is given by \(\frac{a}{1-r}\) when 0<r<1)
= \(\frac{u^2}{g(1-f^2)}\)
we know that--- \(v^2\) = \(u^2\)+2as ( where v= final velocity u= initial a= acceleration acting on body t= time s= distance covered)
when the body is thrown upward with initial velocity = u
its final velocity becomes zero when it reaches its highest point so v=0
acceleration acting on the body in the downward direction = -g m/s^2
\(0^2\)= \(u^2\)-2gs
s = \(\frac{u^2}{2g}\)
now it comes down so total distance = \(\frac{2u^2}{2g}\) (distance going up+distance coming down)
again it rebounds and goes upward with velocity = uf
total distance covered going up and coming down = \(\frac{2u^2f^2}{2g}\)
again it rebounds and goes upward with velocity = u\(f^2\) (we can find distance covered each time by above process)
this process continues till infinity
total distance covered during this process = \(\frac{2u^2}{2g}\)+\(\frac{2u^2f^2}{2g}\)+\(\frac{2u^2f^4}{2g}\)+............till infinity
total distance = \(\frac{u^2}{g}\)(1+\(f^2\)+\(f^4\)+\(f^6\)+............∞) (we see this forms geometric progression)
total distance = \(\frac{u^2}{g}\)(\(\frac{1}{1-f^2}\))....(sum of infinite GP with first term a and common ratio r is given by \(\frac{a}{1-r}\) when 0<r<1)
= \(\frac{u^2}{g(1-f^2)}\)
10 Apr 2026, 03:52
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Can someone explain it in an easier way so that even if I don't know some physics equation , I am still able to solve it.
