Question

An object is thrown up the velocity of 20 metre per second find
1) the maximum height attained by it.
2) the total time taken by it to return back take G is equal to 10 metre per second square.

Answer 1

1. maximum height formula=u^2/g

so h=20^2/10

h=400/10=40 m

2. time of flight formula=2u/g

so t=2*20/10=4 secs

total time=4*2=8 secs

hope this helps

Answer 2

Given that,

Initial speed of the object, u = 20 m/s

To find,

1. The maximum height attained by it.

2 The total time taken by it to return back.

Solution,

(a) At maximum height, the final speed of the object is equal to 0. Now using the conservation of energy as :

$u=\sqrt{2gh} \\\\h=\dfrac{u^2}{2g}\\\\h=\dfrac{(20)^2}{2\times 10}\\\\h=20\ m$

(b) Now using first equation of motion as :

$v=u+at$

Here, a = -g

$v=u-gt\\\\u=gt\\\\t=\dfrac{u}{g}\\\\t=\dfrac{20}{10}\\\\t=2\ s$

So, the total time taken by it to return back is 2 s + 2 s = 4 s.

Learn more,

Kinematics

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