Question

If body is projected at an angle 30⁰ with horizontal with velocity 440 cm/s the time flight is -----

a) 0.4 sec     b) 0.550 sec    
  c) 0.448 sec      d) 0.5 sec​

Answer 1

Given :-

Angle of projection, θ = 30°

Initial velocity, u = 440 cm/s

To Find :-

Time of flight, T

Solution :-

We know :-

$\bf T = \dfrac{2 \: u \: sin \theta}{g}$

Here :-

u = 440 cm/s

ㅤ= 440/100

ㅤ= 4.4 m/s

sin θ = 1/2

g = 9.8 m/s^2

Substitute the values :-

$\sf \implies T =\dfrac{2 \times 4.4 \times \dfrac{1}{2} }{9.8}$

$\sf \implies T = \dfrac{4.4}{9.8}$

$\sf \implies T = 0.448$

Time of flight = 0.448 seconds

Answer 2

Answer:

Given :-

• A body is projected at an angle of 30° with horizontal with velocity 440 cm/s.

To Find :-

• What is the time flight.

Formula Used :-

$\clubsuit$ Time Taken Formula :

$\longmapsto \sf\boxed{\bold{\pink{Time\: taken\: (t) =\: \dfrac{2vsin\theta}{g}}}}$

where,

• v = Velocity
• g = Acceleration due to gravity

Solution :-

First, we have to change cm/s into m/s :

$\implies \sf Velocity =\: 440\: cm/s$

$\implies \sf Velocity =\: \dfrac{440}{100}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: cm/s =\: \dfrac{1}{100}\: m/s}}\bigg\rgroup$

$\implies \sf Velocity =\: \dfrac{44\cancel{0}}{10\cancel{0}}\: m/s$

$\implies \sf Velocity =\: \dfrac{44}{10}\: m/s$

$\implies \sf\bold{\green{Velocity =\: 4.4\: m/s}}$

Now, we have to find the time of flight :

${\normalsize{\bold{\purple{\underline{\bigstar\: Time\: of\: flight\: :-}}}}}$

Given :

$\bigstar$ Velocity = 4.4 m/s

$\bigstar$ Acceleration due to gravity = 9.8 m/s²

$\bigstar$ Angle of projection = 30°

According to the question by using the formula we get,

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{2 \times 4.4 \times sin30^{\circ}}{9.8}$

As we know that, [ sin30° = 1/2 ]

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{\cancel{2} \times 4.4 \times \dfrac{1}{\cancel{2}}}{9.8}$

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{4.4}{9.8}$

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{\dfrac{44}{10}}{\dfrac{98}{10}}$

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{44}{10} \times \dfrac{10}{98}$

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{44\cancel{0}}{98\cancel{0}}$

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{\cancel{44}}{\cancel{98}}$

$\longrightarrow \sf Time\: of\: flight =\: \dfrac{22}{49}$

$\longrightarrow \sf\bold{\red{Time\: of\: flight =\: 0.448\: seconds}}$

$\therefore$ The time of flight is 0.448 seconds.

Hence, the correct options is option no (c) 0.448 seconds.

IMPORTANT FORMULA :-

$\clubsuit$ Horizontal Range Formula :

$\longmapsto \sf\boxed{\bold{\pink{Horizontal\: Range\: (R) =\: \dfrac{v^2sin2\theta}{g}}}}$

$\clubsuit$ Maximum Height Formula :

$\longrightarrow \sf\boxed{\bold{\pink{Maximum\: Height =\: \dfrac{v^2sin^2\theta}{2g}}}}$

where,

• v = Velocity
• g = Acceleration due to gravity