Question

a stone is thrown horizintally wih a velocity of 10m/s .Find the radius of curvature of its trajectory a the end of 3 sec after the motion begins
a)100 root 10
b) 50 root 10
c) 1/50 root 10
d) 1/100 root 10

Answer 1

Object is projected horizontally  with speed 10 m/s

Now after 3 s it will gain vertical speed

so

$v_y = gt = 3*10 = 30 m/s$

now we know that the angle that it will make with the horizontal is given as

$tan\theta = \frac{30}{10}$

$\theta = 71.56 degree$

now the acceleration perpendicular to its net speed is given as

$a = gcos\theta$

$a = 10*cos71.56$

now the radius is given by

$R = \frac{v^2}{a}$

$R = \frac{10^2 + 30^2}{10cos71.56}$

$R = 100*\sqrt10$

so correct answer is option A

Answer 2

Answer:

Explanation:s per given condition of motion the velocity in x direction is v  

x

​  

=10m/s

The velocity in y direction v  

y

​  

=gt=10∗3=30m/s

Resultant velocity v  

r =  

((v  

x

​  

)  

2

+(v  

y

​  

)  

2

)

​  

=  

(10  

2

+30  

2

)

​  

=10  

(10)

​

tan thita =3 thita =71.56

a =gcosthita

a=10*cos71.56

r=v^2/a =10^2+30^2/10*cos71.56= 100$\sqrt{10}$