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Given that a particle is thrown with angle of projection theta over a triangle DeltaACB from one of its end A of the horizontal base AB aligned along X-axis and it finally falls at the other end Bof the base,grazing the vertex C(x,y)
Let u be the velocity of projection, T be the time of flight, R=AB be the horizontal range and t be the time taken by the particle to reach at C (x,y)
The horizontal component of velocity of projection ->ucostheta
The vertical component of velocity of projection ->usintheta
Considering motion under gravity without any air resistance we can write
y=usinthetat-1/2 g t^2.....[1]
x=ucosthetat...................[2]
combining [1] and [2] we get
y=usinthetaxxx/(ucostheta)-1/2 xxgxxx^2/(u^2cos^2theta)
=>y=usinthetaxxx/(ucostheta)-1/2 xxgxxx^2/u^2xxsec^2theta
=>color(blue)(y/x=tantheta-((gsec^2theta)/(2u^2))x........[3])
Now during time of flight T the vertical displacement is zero
So
0=usinthetaT-1/2 g T^2
=>T=(2usintheta)/g
Hence horizontal displacement during time of flight i.e. range is given by
R=ucosthetaxxT=ucosthetaxx(2usintheta)/g=(u^2sin2theta)/g
=>R=(2u^2tantheta)/(g(1+tan^2theta))
=>R=(2u^2tantheta)/(gsec^2theta)
=>color(blue)((gsec^2theta)/(2u^2)=tantheta/R......[4])
Combining [3] and [4] we get
y/x=tantheta-1/2 xx(gx)/u^2xxsec^2theta
=>y/x=tantheta-(xtantheta)/R
=>tanalpha=tantheta-(xtantheta)/R [since color(red)(y/x=tanalpha) from figure]
So tantheta=tanalphaxx(R/(R-x))
=>tantheta=tanalphaxx((R-x+x)/(R-x))
=>tantheta=tanalphaxx(1+x/(R-x))
=>tantheta=tanalpha+(xtanalpha)/(R-x)
=>tantheta=tanalpha+y/(R-x) [putting color(red)(xtanalpha=y)]
Finally we have from figure color(magenta)(y/(R-x)=tanbeta)
Hence we get our required relation
color(green)(tantheta=tanalpha+tanbeta)
