Question

11. A body is projected vertically up with a velocity of g ms' from the top of a tower of height
$height \: of \: tower = \frac{ {u}^{2} }{2g}$


The maximum height reached by the body from ground is k(g), where k =

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Answer 1

Answer:

5m

Explanation:

Initial velocity of the body, u=10 m/s (Taking upward motion to be positive)

Acceleration of the body, a=g=−10 m/s

2

At maximum height, the final velocity of the body is zero, i.e. v=0

Using: v

2

−u

2

=2gS

∴ 0−10

2

=2(−10)H

max

⟹ H

max

=

2×10

10

2

=5 m