Question

a body is projected up with an initial velocity of 10m/s. it will return to the stating point after

Answer 1

v=u-gt
0=u-gt(at maximum height)
t=u/g
t=10/10
t=1 sec to reach maximum height
total time taken to return to the stating point is 2 sec

Answer 2

that is the query of movement beneathneath gravity , it follows movement beneathneath regular accleration.

Equations

Acceleration is regular over the time interval
v = u + at
v² = u² + 2as
s = ut + ½at²

in which, s = displacement; u = preliminary speed; v = very last speed; a = acceleration; t = time of movement. These equations are referred as SUVAT equations in which SUVAT stands for displacement (s), preliminary speed (u), very last speed (v), acceleration (a) and time (T)


The frame is thrown up with preliminary speed v = 10m/ s.

We must discover the time and then it'll be returning returned to the authentic region of throwing.
We can use the formula v =u + at for locating the proper answer

v=u-gt
0=u-gt(at most height)
t=u/g
t=10/10
t=1 sec to attain most height
overall time taken to go back to the pointing out factor is two sec

= 2
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