Question

The horizontal range of a projectile is R and the maximum height at tained by it is H. A strong windnow begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be:

Answer 1

Answer:

Let the angle of projection be θ and speed of projectile be u.

Initially, horizontal component of velocity u

x

=ucosθ

Acceleration in horizontal direction a=

2

g

So, horizontal range of projectile R

′

=u

x

T+

2

1

aT

2

where T=

g

2usinθ

is the time of flight.

Or R

′

=(ucosθ)

g

2usinθ

+

4

g

×(

g

2usinθ

)

2

Or R

′

=

g

u

2

sin2θ

+2

2g

u

2

sin

2

θ

Using R=

g

u

2

sin2θ

and H=

2g

u

2

sin

2

θ

⟹ R

′

=R+2H

Answer 2

Answer:

Let the angle of projection be θ and speed of projectile be u.

Initially, horizontal component of velocity u

x

=ucosθ

Acceleration in horizontal direction a=

2

g

So, horizontal range of projectile R

′

=u

x

T+

2

1

aT

2

where T=

g

2usinθ

is the time of flight.

Or R

′

=(ucosθ)

g

2usinθ

+

4

g

×(

g

2usinθ

)

2

Or R

′

=

g

u

2

sin2θ

+2

2g

u

2

sin

2

θ

Using R=

g

u

2

sin2θ

and H=

2g

u

2

sin

2

θ

⟹ R

′

=R+2H