Question

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Answer 1

QUESTION:-

A body is thrown up with a velocity of 45 m/s.Its disp. after 5 second

EXPLANATION:-

Let first calculate time taken by the object to reach top:-

Using 1st equations:-

v=u+gt

0=45-(10t)

t=4.5 s

So the time taken to reach maximum height is 4.5 s

Now let's calculate the maximum distance of obj from ground using 2nd eq,

S=ut+1/2 at²

S=45×4.5+(1/2)(-10)(4.5)²

S=202.5 -5×(20.5)

S=202.5-101.5

S=101.5 m

So maximum height is 101 metres

Now let's calculate the distance travelled in 5 second

S=ut+1/2at²

S=45×4-1/2(10)4²

S=180-100

S=80 metres

So distance is 80 metres

Now disp.=101.25-80

==>21.25

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Answer 2

QUESTION:-

A body is thrown up with a velocity of 45 m/s.Its disp. after 5 second

EXPLANATION:-

Let first calculate time taken by the object to reach top:-

Using 1st equations:-

v=u+gt

0=45-(10t)

t=4.5 s

So the time taken to reach maximum height is 4.5 s

Now let's calculate the maximum distance of obj from ground using 2nd eq,

S=ut+1/2 at²

S=45×4.5+(1/2)(-10)(4.5)²

S=202.5 -5×(20.5)

S=202.5-101.5

S=101.5 m

So maximum height is 101 metres

Now let's calculate the distance travelled in 5 second

S=ut+1/2at²

S=45×4-1/2(10)4²

S=180-100

S=80 metres

So distance is 80 metres

Now disp.=101.25-80

==>21.25