Question

a particle is thrown such that it returns back to ground after 9s Find the distance travelled in 5th second

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Answer 1

Answer:

s = ut + (1/2)at2 As the particle comes to the same point as 9s where it was at 5s. The net displacement at 4s is zero. ← Prev QuestionNextRead more on Sarthaks.com - https://www.sarthaks.com/126159/particle-thrown-upwards-attains-height-seconds-again-after-comes-speed-particle-height

Answer 2

Explanation:

Given that,

Given that,Velocity v=um/s

Given that,Velocity v=um/sTime taken to reach at maximum height t

Given that,Velocity v=um/sTime taken to reach at maximum height t 1

Given that,Velocity v=um/sTime taken to reach at maximum height t 1

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5s

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9s

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5s

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4s

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4 t=0.5s

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4 t=0.5sSo, the velocity at height h

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4 t=0.5sSo, the velocity at height h v=u−gt

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4 t=0.5sSo, the velocity at height h v=u−gt 0=u−9.8×0.5

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4 t=0.5sSo, the velocity at height h v=u−gt 0=u−9.8×0.5 u=4.9m/s

Given that,Velocity v=um/sTime taken to reach at maximum height t 1 =5sNow,Total time taken to reach same point t 2 =9sParticle to reach its maximum height will be half of total time =4.5sNow, the particle returns to the same point in time t=9−5=4sNow, the time at height is t=4.5−4 t=0.5sSo, the velocity at height h v=u−gt 0=u−9.8×0.5 u=4.9m/sHence, the speed of the particle at height is 4.9 m/s

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