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+

Answer 1

Let the velocity of the plane be V m/s

horizontal component of velocity =

V cos 30°= 200

v x √3/2 = 200

v = 400/√3

=231 km/hr

vertical component of velocity is

V sin 30 = v/2=115.47 km/h

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....