With what speed should a body be thrown up wards is that the distance traversed

Question

With what speed should a body be thrown up wards is that the distance traversed in 5^th second and 6^th second are equal? 58.4 m/s 49 m/s squareroot 78 m/s 98 m/s Water drops fall at regular intervals from a tap 5 m above the found. The third drop is having the top at the instant that first drop the ground. How far above the ground is second drop at that instant 1.25 m 2.50 m 3.75 m 4.00 m The velocity required by a body moving with uniform acceleration is 30 ms^-1 in 2 seconds and 60 ms^-1 in 4 seconds. The initials velocity zero 2 ms^-1 4 ms^-1 10 ms^-1 A rocked is launched upwards from the earth is surface whose velocity time graphs in figures. Than maximum height other by that rocket is 1 km 10 km 100 km 60 km In above question the retardation of is In above question the rockets goes up and comes down on the following parts respectively. From the following velocity time graph of a body the distance traveled by the body and its displacement during 5 seconds in meters will be: 75, 75 110, 70 110, 110 110, 40

Explanation / Answer

In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s

In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s

In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s

In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s

In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s

In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s In the given question it is given that the body is thrown upward , taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight. Here 5th second means between t = 4 & t= 5 Similarly 6th second means  between t=5 & t=6 Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5 Thus body is at same height at t=4 and t=6 Hence total time of flight is 10 second= 2u/g thus u = 50m/s

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