Physics - Class 11
Draw this answer on x-axis and y-axis for better understanding.
Let your tent is at a point, we say it O (0,0), Joe's tent is at Point J (in the first quadrant) and Karl's tent is at Point K (in the fourth quadrant). So now acc. to the ques. line joining OJ will make angle of 37.0 and line joining OK will make angle of 23.0 with the x-axis.
Given, OJ = 21m. and OK = 32m.
Therefore; vector OJ = (21xcos23)i - (21xsin23)j
and vector OK = (32xcos37)i + (32xsin37)j
From vector triangle,
OJ + JK = OK => JK = OK - OJ
vector JK = (32xcos37 - 21xcos23)i + (32xsin37 + 21xsin23)j
|JK|^2 = [(32xcos37 - 21xcos23)^2 + (32xsin37 + 21xsin23)^2] m^2
Jk = 28.16m
So, the distance between Joe and Karl tent is = 28.16m or 28.20m
Hello
The equation g÷ 2u^2cos^2theta is used to solve Projectile Motion Problems to make make that equation represents Equation of Parabola
Check the Derivation here
Let projectile is thrown with a initial velocity u at an angle theta there with horizontal.
U=uxi+yuk
Ux=ucos theta
Uy=usin theta
Neglecting air resistance
Along x direction
S=ut+1/2at^2
Distance covered by projectile = velocity time
t=X(x)/ucos theta ----------(1)
Along y direction
y=usin theta × t-1/2gt^2 ---------(2)
Substituting (1) in (2)
y= usin theta × x/ucos theta - 1/2 g (x/ucos theta)^2
y=x tan theta - (g/2u^2cos^2theta
Let tan theta =A
-g/2u^2cos^2 theta =B
Y= Ax+Bx^2--------(Parabola)
This equation represents equation of parabola.
Hence path of projectile(Trajectory) is a parabola
Hope this Helps
All the best
There are seven physical fundamental quantities. Its been listed in the Measurement chapter of Physics 11th standard book as well.
I will enlist them down below with their SI units..
Best wishes. Thank you.
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