Question

A stone is thrown horizontally under gravity with speed of 10m sec find radius of curvature at the end of 3 sec

Answer 1

Answer:

Explanation:

As per given condition of motion the velocity in x direction is v x=10 m/s

The velocity in y direction  v y=gt=10∗3=30 m/s

Resultant velocity v r=√((v x)2+(v y)2)=√(102+302)=10√(10)

So the answer will be 10√(10)

Answer 2

Answer:

$100\sqrt{10}$

Explanation:

Object is projected horizontally  with speed 10 m/s

after 3 s it will gain vertical speed

$v_{y} = 0 + (g)(3)$           [as $u_{y} =0$ & $a_{y} =g$ ]

$v_{y} = 30$

$v_{x} = u_{x} =10$               [ $a_{x} =0$ ]

v after 3 sec = $\sqrt{v_{y}^2 + v_{x} ^2} =\sqrt{30^2 + 10^2} =\sqrt{1000}$

the acceleration perpendicular to its net velocity is given as

$a_{t} = mg sin\alpha$

as $tan\alpha = \frac{v_{y}}{v_{x} } =\frac{30}{10} =3$

perpendicular = 3

base = 1

hypotaneous = $\sqrt{3^{2} + 1^2} =\sqrt{10}$

$sin\alpha =\frac{1}{\sqrt{10} }$

$a_{t} =\frac{mv^2}{r}$

 $r=\frac{mv^2}{a_{t} }$

   $= \frac{(m)(1000)}{(m)(g)(sin\alpha) }$

   $= \frac{(m)(1000)}{(m)(10)(\frac{1}{\sqrt{10} } ) }$

   $= 100\sqrt{10}$ m

Therefore, radius of curvature at the end of 3 sec is = $= 100\sqrt{10}$

Please mark as brainliest