Question

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle. At the bottom of the dive, the speed of the airplane is 320 m/s. What is the smallest radius allowable for the vertical circle if the pilot's apparent weight is not to exceed 7.0 times his true weight

Answer 1

Answer:

1.74 km

Explanation:

We are given that

Speed of airplane,v=320 m/s

We have to find the smallest radius allowable for the vertical circle if the pilot's apparent weight is not exceed 7.0 times his true weight.

Let true weight =w

Apparent weight=7w

According to question

$7w=w+\frac{mv^2}{r}$

$7w-w=\frac{mv^2}{r}$

We know that

$m=\frac{w}{g}$

$6w=\frac{wv^2}{rg}$

$6=\frac{v^2}{rg}$

$r=\frac{v^2}{6g}$

Where $g=9.8 m/s^2$

$r=\frac{(320)^2}{6\times 9.8}= 1741.5 m$

$r=\frac{1741.5}{1000}=1.74 km$

Where 1 km=1000 m