Question

An air craft executes a circular loop at a speed of 200 ms⁻¹ with its wings banked at 45°. The radius of the loop is

Answer 1

Answer:

Using the relation for the radius (r) of loop.

tanθ=

rg

v

2

or, tan12

o

=

r×10

(150)

2

or, r=

0.2125

2250

=10.6×10

3

m=10.6km

Answer 2

Explanation:

GIVEN :-

• Velocity of air craft , v = 200 m/s.

• Angle of Wings banking , ∅ = 45°.

• Acceleration due to gravity , g = 10 m/s².

TO FIND :-

• The radius of the loop.

SOLUTION :-

$\: \underline{ \boldsymbol{we \: know \: the \: relation,}}$

$:\implies \displaystyle \sf \tan \theta = \dfrac{v^{2}}{rg}$

$:\implies \displaystyle \sf \tan 45 ^{ \circ} = \dfrac{200^{2}}{r \times 10}$

$\\ :\implies \displaystyle \sf 1 = \dfrac{40000}{r \times 10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \: \tan45^{ \circ} = 1\bigg \rgroup$

$:\implies \displaystyle \sf r = \dfrac{40000}{10}$

$:\implies \underline{ \boxed{ \displaystyle \sf r \: = 4000 \: cm. }}$

___________________

$\because\displaystyle \sf 1 \: km = 1000 \: cm.$

$\dashrightarrow \: \displaystyle \sf r \: = 4000 \: cm.$

$\dashrightarrow \: \displaystyle \sf r = \dfrac{4000}{1000} \: km$

$\dashrightarrow \: \displaystyle \sf r = 4 \: km \: \: \: \: \: \: \: \bigg \lgroup radius \: of \: lo \ op \bigg \rgroup$

$\therefore \: \underline {\displaystyle \sf radius \ of \ the \ lo \ op \ is \ 4 km.}$