A body is thrown up with an initial velocity of 10-m-s-1- It reaches a maximum height- h- Find the value of h- nbsp- -Take acceleration due to gravity- g-10-m-s-2- 10-m-15-m-5-m-20-m-

The correct option is C 5 m Given, initial velocity, u=10 m s^ 1 Using equation, v^2 = u^2 nbsp;+ 2as Here, a = g nbsp; [∵ acceleration due to gravity is ...

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Standard IX

Physics

Free Fall

A body is thr...

Question

A body is thrown up with an initial velocity of $10 m s^{- 1}$ . It reaches a maximum height, $h$ . Find the value of $h$ . [Take acceleration due to gravity, $g = 10 m s^{- 2}$ ]

$10 m$

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$15 m$

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$5 m$

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$20 m$

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Solution

The correct option is C
$5 m$

Given,
initial velocity, $u = 10 m s^{- 1}$
Using equation,
$v^{2}$ = $u^{2} + 2 a s$
Here, $a = - g$ [ $∵$ acceleration due to gravity is acting opposite to the direction of motion]
$s = h$
At highest point,
Final velocity, $v = 0$
$\Rightarrow 0 = u^{2} - 2 g h$
$\Rightarrow h = \frac{u^{2}}{2 g}$
$= \frac{10 \times 10}{2 \times 10} = 5 m$

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A body is thrown up with an initial velocity of 10 m s−1. It reaches a maximum height, h. Find the value of h. [Take acceleration due to gravity, g=10 m s−2]

The correct option is C 5 m Given, initial velocity, u=10 m s−1 Using equation, v2 = u2+2as Here, a=−g [∵ acceleration due to gravity is acting opposite to the direction of motion] s=h At highest point, Final velocity, v=0 ⇒0=u2−2gh ⇒h=u22g =10×102×10=5 m

A body is thrown up with an initial velocity of $10 m s^{- 1}$ . It reaches a maximum height, $h$ . Find the value of $h$ . [Take acceleration due to gravity, $g = 10 m s^{- 2}$ ]