Question

A particle is thrown horizontally with speed 20 ms–1 from the top of a high tower. What is the radius of curvature of the particle 2 s later? [Neglect air friction and take g = 10 ms–2]

Answer 1

Answer:

45

Explanation:

v

x

=u

x

=20

using: v

2

−u

2

=2as to find v

y

v

y

=

2gh

v

y

=20

tanθ=

v

x

v

y

tanθ=20/20=1

θ=45

o

Answer 2

Answer:

The correct answer is equal to $80\sqrt{2}m$.

Explanation:

Given:

A Particle is thrown horizontally with a speed of 20m/sec.

V denoted as a particle thrown horizontally.

u denoted as the initial speed of a particle in the vertical direction.

t denoted as time.

a denoted as acceleration due to gravity.

The Horizontal component of the velocity of the particle after 2sec,

$V_{h}=V\\V_{h}=20m/sec$

The verticle component of the velocity of the particle after 2sec,

$V_{v}= u+at$

$V_{v}$= 0+(-10)×2

$V_{v}=-20m/sec$

Resultant of velocity $V_{v}$ and $V_{h}$,

$V_{r}=\sqrt{V_{v}^{2} +V_{h}^{2} }$

$V_{r}=\sqrt{(-20)^{2} +20^{2} }\\V_{r}=20\sqrt{2}m/sec$

The angle between $V_{r}$ and $V_{h}$,

TanФ $=\frac{V_{v}}{V_{h}}$

         $=\frac{-20}{20 }$

Ф= -45°    

The component of acceleration is normal to velocity,

$a_{n}=a$×cosФ

    = 10×cos(-45°)

    = $5\sqrt{2}m/sec^{2}$

The radius of curvature after 2sec,

$R=\frac{v_{r}^{2} }{a_{n} } \\R=\frac{(20\sqrt{2}) ^{2} }{5\sqrt{2} }$

R = $80\sqrt{2}m$

So, the radius of curvature is equal to  $80\sqrt{2}m$.