Question

A body is projected at 60 degrees with the horizontal with velocity of 10 root 3 meters per second.The velocity of the projectile when it moves perpendicular to its initial direction is

Answer 1

Answer:

Velocity along x axis remains constant.

Let velocity along y axis at that instant be v.

So, according to the question

tan30

∘

=

10cos60

∘

v

=

5

v

3

1

=

5

v

v=

3

5

speed=

5

2

+(

3

5

)

2

=

25+

3

25

=

3

100

=

3

10

m/s.

Answer 2

The velocity of the projectile when it moves perpendicular to its initial direction is 10 m/s.

A body is projected at 60° with the horizontal with velocity of 10√3 m/s.

We have to find the velocity of the projectile when it moves perpendicular to its initial direction.

The body is projected with a speed of 10√3 m/s at an angle 60° with horizontal.

∴ Initial velocity of the projectile is given by,

$\vec u=(ucos\theta)\hat{i}+(usin\theta)\hat{j}\\\\=(10\sqrt{3}cos60^{\circ})\hat i+(10\sqrt{3}sin60^{\circ})\hat j\\\\=(5\sqrt{3})\hat i+(15)\hat j$

Let after time t, velocity of the projectile becomes perpendicular to its initial direction.

∴ Velocity of the projectile after time t is given by,

$\vec v=(ucos\theta)\hat i+(usin\theta-gt)\hat j\\\\=(10\sqrt{3}cos60^{\circ})\hat i+(10\sqrt{3}sin60^{\circ}-gt)\hat j\\\\=(5\sqrt{3})\hat i+(15-gt)\hat j$

Now, both the velocities are perpendicular to each other.

$\vec u.\vec v=0\\\\\implies((5\sqrt{3})\hat i+(15)\hat j).((5\sqrt{3})\hat i+(15-gt)\hat j)=0\\\\\implies (5\sqrt{3})^2+15(15-gt)=0\\\\\implies75+225-15\times10t=0\\\\\implies t=2s$

Now the velocity of the projectile will be..

$\vec v=(5\sqrt{3})\hat i+(15-10\times2)\hat j\\\\=(5\sqrt{3})\hat i+5\hat j$

∴ The magnitude of the velocity is

$|\vec v|=\sqrt{(5\sqrt{3})^2+(5)^2}=10$

Therefore the velocity of the projectile when it moves perpendicular to its initial direction is 10 m/s.

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