UPSEE Solved Model Paper 7

Given ：
Nozzle velocity of the defense cannon, u = 85 m/s
Distance between the defense cannon and the enemy ship, d = 575 m
The cannon will execute a projectile motion and hit the enemy ship
Thus, Distance between the defense cannon and the enemy ship = Horizontal range of the cannon
We know that the horizontal range is given by
R =

u2sin(2θ)

g

... (i)
⇒ sin (2θ) =

Rg

u2

⇒ θ =

1

2

×sin−1(

Rg

u2

)
⇒ θ =

1

2

×sin−1[

575m×9.8m∕s2

(85m∕s)2

]
⇒ θ =

1

2

×sin−1(0.78)
Using sin−1 0.78 = 51.26° , we get
θ =

1

2

× 51.26°
⇒ θ = 25.63°
Now, re-writing eq (i) as
R =

u2sin(2θ)

g

=

2u2sinθcosθ

g

[as sin (2θ) = 2 sin θ cos θ]
If we change θ to "90° - θ" we have
R' =

2u2sin(90°−θ)cos(90°−θ)

g

⇒ R' =

2u2cosθsinθ

g

=

u2sin(2θ)

g

= R
i.e., range remains the same
Therefore, the corresponding angles for the cannon are 25.63° and 90° - 25.63° = 64.37°
The commandant of the fort can elevate the gun to either if the two angles and hit the enemy ship
Hence, option 'C' is correct