A stone projected from ground with certain speed at an angle #alpha#with horizontal attains maximum height #h_1#when it is projected with same speed at an angle#alpha# with vertical attains height #h_2#.What is the horizontal range of projectile?

1 Answer
Aug 17, 2017

drawn

Deduction of formulas

Let the velocity of projection of the stone be u with angle of projection #theta# with the horizontal direction.
The vertical component of the velocity of projection is #usintheta# and the horizontal component is #ucostheta#

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical displacement h will be zero. So applying the equation of motion under gravity we can write

# h=usintheta xxT+1/2gT^2 #

#=>0=usinthetaT-1/2xxgxxT^2# where #g="acceleration due to gravity"#

#:.T=(2usintheta)/g#
The horizontal displacement or range during this T sec is

#R=ucosthetaxxT#

#=>color(red)(R=(u^2sin(2theta))/g.......[1])#

If H be the maximum height attained by the stone we can write

#0^2=u^2sin^2theta-2gH#

#color(red)(=>H=(u^2sin^2theta)/(2g)....[2])#

CALCULATION USING ABOVE RELATIONS

For 1st case

when #theta=alpha,H=h_1#

So horizontal range

#R_1=(u^2sin(2alpha))/g.....[3]#

and # h_1=(u^2sin^2alpha)/(2g)....[4]#

For 2nd case

when #theta=90-alpha,H=h_2#

So horizontal range

#R_2=(u^2sin(2(90-alpha)))/g#

#=>R_2=(u^2sin(180-2alpha))/g#

#=>R_2=(u^2sin(2alpha))/g....[5]#

and # h_2=(u^2sin^2(90-alpha))/(2g)#

#=>h_2=(u^2cos^2(alpha))/(2g)....[6]#

In both the cases the range is same.So let #R_1=R_2=R#

Hence range

#R=(u^2sin(2alpha))/g#

Now multiplying [4] and [6] we get

#h_1h_2=(u^4sin^2alphacos^2alpha)/(2g)^2#

#=>h_1h_2=(u^2xx2sinalphacosalpha)^2/(4g)^2#

#=>h_1h_2=(u^2xxsin2alpha)^2/(4g)^2#

#=>sqrt(h_1h_2)=(u^2xxsin2alpha)/(4g)#

#=>sqrt(h_1h_2)=R/4#

#color(red)(=>R=4sqrt(h_1h_2))#