Question

An Aeroplane takes off at angle of
30 to the horizontal . f the component
of its velocity among horizontal is
250 km lhes .What is the actual
actual velocity . Also find the vertical
component of the velocity​

Answer 1

An aeroplane takes off at an angle 30° to the horizontal as shown in figure.

given, f is the component of its velocity along horizontal is 250 km/hr

see figure,

we assume , v is the velocity of aeroplane makes an angle 30° to the horizontal.

so, horizontal component of v = vcos30° = f = 250 km/hr [ from question]

or, v × (√3/2) = 250

or, v = 250 × 2/√3 = 500/√3 km/hr ≈ 288.675 km/h

hence, actual velocity of aeroplane is 288.67 km/h

now vertical component of velocity = vsin30° = (500/√3) × 1/2 = 250/√3

now vertical component of velocity = vsin30° = (500/√3) × 1/2 = 250/√3 ≈ 144.33 km/hr

Answer 2

$\huge {\bf {\underline {Question}}}$

An aeroplane takes off at an angle of 30 degree to the horizontal. if the component of its velocity along teh horizontal is 250 km/hr, what is the actual velocity. find also its vertical component.

$\huge {\bf {\underline {Answer}}}$

Let the velocity of the plane be V m/s.

Horizontal component of velocity =

$V cos 30° = 250 km/h \\ \implies V \frac {\sqrt {3}}{2} = 250 \\ \implies V = \frac {250 \times 2}{\sqrt {3}} \\ \implies V = \frac {500}{\sqrt {3}} \\ \implies V = \frac {500}{1.732} \\ \implies V = 288.7 \: km/h$

Vertical component of velocity =

$V sin 30° = \frac {V}{2} = \frac {288.7}{2} = 144.35\:km/h$

Hope it helps dear....