A body is projected with initial velocity of 40 m/s and at an angle of 30°with the horizontal. Calculate (i) time of flight and (i i) horizontal range of projectile.(assume g= 10 m/s²) reply me
Answers
Answer:
Answer:
Given :-
A body is projected with initial velocity of 40 m/s and at an angle of 30° with the horizontal. (assume g = 10 m/s²).
To Find :-
Time of flight
Horizontal range of projectile.
Formula Used :-
\clubsuit♣ Time Taken :
\longmapsto\sf\boxed{\bold{\pink{Time\: taken\: (t) =\: \dfrac{2vsin\theta}{g}}}}⟼Timetaken(t)=g2vsinθ
\clubsuit♣ Horizontal range :
\longmapsto\sf\boxed{\bold{\pink{Horizontal\: range\: (R) =\: \dfrac{{v}^{2}sin2\theta}{g}}}}⟼Horizontalrange(R)=gv2sin2θ
where,
v = Velocity
g = Acceleration due to gravity
Solution :-
{\small{\bold{\purple{\underline{i)\: Time\: of\: flight\: :-}}}}}i)Timeofflight:−
Given :
Velocity = 40 m/s
Acceleration due to gravity = 10 m/s²
Angle of projection = 30°
According to the question by using the formula we get,
\begin{gathered}\implies \sf Time\: of\: flight =\: \dfrac{2 \times 40 \times sin30^{\circ}}{10}\\\end{gathered}⟹Timeofflight=102×40×sin30∘
As we know that, [ sin30° = 1/2 ]
\implies \sf Time\: of\: flight =\: \dfrac{\cancel{80} \times \dfrac{1}{\cancel{2}}}{10}⟹Timeofflight=1080×21
\begin{gathered}\implies \sf Time\: of\: flight =\: \dfrac{4\cancel{0}}{1\cancel{0}}\\\end{gathered}⟹Timeofflight=1040
\implies \sf Time\: of\: flight =\: \dfrac{4}{1}⟹Timeofflight=14
\implies\sf\bold{\red{Time\: of\: flight =\: 4\: seconds}}⟹Timeofflight=4seconds
\therefore∴ The time of flight is 4 seconds.
\rule{150}{2}
{\small{\bold{\purple{\underline{ii)\: Horizontal\: range\: of\: projectile\: :-}}}}}ii)Horizontalrangeofprojectile:−
Given :
Velocity = 40 m/s
Acceleration due to gravity = 10 m/s²
Angle of projection = 30°
According to the question by using the formula we get,
\begin{gathered}\implies \sf Horizontal\: Range =\: \dfrac{{(40)}^{2} \times sin2 \times 30^{\circ}}{10}\\\end{gathered}⟹HorizontalRange=10(40)2×sin2×30∘
\begin{gathered}\implies \sf Horizontal\: Range =\: \dfrac{1600 \times sin60^{\circ}}{10}\\\end{gathered}⟹HorizontalRange=101600×sin60∘
As we know that, [ sin60° = √3/2 ]
\implies \sf Horizontal\: Range\: =\: \dfrac{\cancel{1600} \times \dfrac{\sqrt{3}}{\cancel{2}}}{10}⟹HorizontalRange=101600×23
\implies \sf Horizontal\: Range =\: \dfrac{80\cancel{0} \times \sqrt{3}}{1\cancel{0}}⟹HorizontalRange=10800×3
\implies \sf Horizontal\: Range =\: \dfrac{80 \times \sqrt{3}}{1}⟹HorizontalRange=180×3
\implies \sf Horizontal\: Ran
Step-by-step explanation:
Correct option is
C
(a)141.4m
(b)20.41m
(c)2.041s
R.E.F image
u=40m/s
θ=30
∘
g=9.81
Time of flight, T=
g
2usinθ
Maximum Height, H=
2g
u
2
sin
2
θ
Horizontal Range, R=
g
u
2
sin2θ
Time taken to reach the
max.height, t =
2
T
4=
2×9.8
(40)
2
(sin30
∘
)
2
=20.408m
=20.41m
R=
9.8
(40)
2
sin2×30
∘
=141.4m
t=
2
I
=
g
usin3θ
=2.041s
Ans =141.4m,20.41m,2.041s (C). this is your ans. I HOPE IT'S HELPFUL MARK ME BRAINLEST PLZ PLEASE THANK YOU