InterviewSolution
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InterviewSolution
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A particle is parojected vertically upwards from grund with initial velocity `u`. a. Find the maximum height `H` the particle will attain and time `T` that it will attain and time `T` that it will take to return to the ground . . b. What is the velocity when the particle returns to the ground? c. What is the displacement and distance travelled by the particle during this time of whole motion. |
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Answer» Cosnider the motion from `A` to `B`: `s=+H` (final point lies above the initial point), initial velocity `=u`, final velocity `v=0`. Let the time taken to go from `A` to `B` be `t_(1)`. Using `v=u-g t`, we get `0=u-g t_(1) rArrt_(1)=(u)/(g)` Using `s=ut-(1)/(2)g t^(2)`, `rArr h=u ((u)/(g))-(1)/(2)g((u)/(g))^(2)=(u^(2))/(2g)` So the maximum height attained is `H=(u^(2))/(2g)`. Cosnsider the return motion from `B` to `A`: `s=H` (final point lies below the initial point) `u=0` (at point `B`, velocty is zero) Let time taken to go from `B` to `A`be `t_(2)`. We have `t_(2)=sqrt((2H)/(g))=sqrt((2)/(g)(u^(2))/(2g))(u)/(g)` Hence `t_(1)` is known as the time of ascent and `t_(2)` is known as the time of descent. We can see that time of ascent=Time of descent`=(u)/(g)` Total time flight `T=t_(1)+t_(2)=(2u)/(g)` Hence, time of flight is the time for which the particle remains in air. Alternative method to find the time of flight: a. Consider the motion from `A` to `B`: `s=0` (initial and final points are same ) Initial velocity `=u`, time taken`=T` Using `s=ut-(1)/(2)g t^(2)`, we have `0=uT-(1)/(2)g t^(2)` `rArr T =(2u)/(g)` b. Magnitude of velocity on returning the fround will be same as that of inital velocity but directionwill be opposite. Proof:Let `v` be the velocity on reaching the ground. Then from previous formulae, we get ` v=- sqrt(2gH)=-sqrt(2g(u^(2))/(2g))rArrv=-u`, hence proved, c. Displacement`=0`, distance travelled `=2H=u^(2)/(g)`. |
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