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 ...

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, $g = 10 m s^{- 2}$ ]

$10 m$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$15 m$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$5 m$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$20 m$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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, $v = 0$
$\Rightarrow 0 = u^{2} - 2 g h$
$\Rightarrow h = \frac{u^{2}}{2 g}$
$= \frac{10 \times 10}{2 \times 10} = 5 m$

Suggest Corrections

Similar questions

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, $g = 10 m s^{- 2}$ ]

A body is thrown vertically up with an initial velocity of 60 m/s. If acceleration due to gravity g = $10 m / s^{2}$ , what is the time taken to reach a height of 100 m. [3 MARKS]

A ball is thrown with an initial velocity of 20 m/s. Find the time of flight and the maximum height reached by the body.
(Assume acceleration due to gravity $g = 10 m / s^{2}$ .

Q. A ball is thrown up with a velocity of

10 m / s

. The maximum height it reaches is:
(Take

g = 10 m / s^{2}

)

Q. A body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at 'h' height above at an interval of 10 s. The height 'h' is (g = 10 m/s^2) (1.) 22 m (2.) 10.2

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, $g = 10 m s^{- 2}$ ]