Question

A body is projected into air with velocity of 19.6 m/S and tita=30 degrees. Find time of flight

Answer 1

I hope it will help you

Answer 2

GIVEN :-

• Velocity of projection is 19.6 m/s
• Angle of projection = 30°

TO FIND :-

• Time of flight of the particle

SOLUTION :-

Angle = 30°

Initial velocity = 19.6 m/s

$\large\rm\bold{\boxed{Time\:of\:flight\:=\:\frac{2usin\theta}{g}}}$

Where ,

• u is initial velocity of particle (or) the velocity the which the particle was projected.
• θ is the angle of projection
• g is acceleration due to gravity

$\large\rm{\rightarrow{Time\:of\:flight\:=\:\frac{2(19.6)[sin(30°)]}{9.8}}}$

$\large\rm{\rightarrow{Time\:of\:flight\:=\:2(2)\times\:\frac{1}{2}}}$

$\large\rm\bold{\boxed{Sin(30°)\:=\:\frac{1}{2}}}$

$\large\rm{\rightarrow{Time\:of\:flight\:=\:2\:sec}}$

∴ The time of flight of the particle is 2 sec

$\large\tt{\underline{\underline{\green{Additional\:Information:-}}}}$

❃ The Horizontal distance (Range ) covered by the particle which is projected with an initial velocity 'u' and angle 'θ' with the horizontal is given by ,

$\large\rm\bold{\boxed{Range\:=\:\frac{u^2Sin2\theta}{g}}}$

❃ The maximum height reached by a particle projected with initial velocity 'u' and angle 'θ' with the horizontal is given by ,

$\large\rm\bold{\boxed{Maximum\:height\:=\:\frac{u^2sin^2\theta}{2g}}}$