-->

Motion In A Plane

Question
CBSEENPH11018596

A stone is projected with velocity (√2gh) from the top of building h meter high. Show that it will fall farthest at a distance (2√2)h.

Solution
Let the stone be projected at angle θ to make the stone fall at farthest point and (0,R) be the coordinates of point where stone reaches the ground. 
 
Here (0, R) lies on the trajectory. Therefore, we have, 
0 space equals space straight h space plus space left parenthesis tanθ right parenthesis straight R space minus space open parentheses fraction numerator straight g over denominator 2 straight u squared end fraction sec squared straight theta close parentheses straight R squared space space space space space space space space... left parenthesis 2 right parenthesis space

For space straight r space to space be space maximum comma space dR over dθ space equals space 0.
So comma space differentiating space equation space left parenthesis 2 right parenthesis space straight w. straight r. straight t space straight theta comma space we space get

0 space equals space 0 plus left parenthesis sec squared straight theta right parenthesis straight R space plus space tanθ dR over dθ

space space space space equals negative fraction numerator straight g over denominator 2 straight u squared end fraction left parenthesis sec squared straight theta right parenthesis 2 straight R dR over dθ space minus space fraction numerator straight g over denominator 2 straight u squared end fraction left parenthesis 2 space sec squared θtanθ right parenthesis straight R squared

Putting space dR over dθ space equals space 0 comma space we space get space

space space space space space space space left parenthesis sec squared straight theta right parenthesis straight R space minus space fraction numerator straight g over denominator 2 straight u squared end fraction left parenthesis 2 space sec squared θtanθ right parenthesis straight R squared space equals space 0
rightwards double arrow space space space space space tan space straight theta space equals space straight u squared over gR equals fraction numerator 2 gh over denominator gR end fraction equals fraction numerator 2 straight h over denominator straight R end fraction space space space space space space space... left parenthesis 3 right parenthesis

Now comma space substituting space value space of space tanθ space in space equation space left parenthesis 2 right parenthesis comma space we space get

space space space space space straight h space plus space straight u squared over gR straight R space minus open square brackets fraction numerator straight g over denominator 2 straight u squared end fraction open parentheses 1 plus fraction numerator straight u to the power of 4 over denominator straight g squared straight R squared end fraction close parentheses close square brackets straight R squared space equals space 0

rightwards double arrow space space straight h space plus straight u squared over straight g minus fraction numerator straight g over denominator 2 straight u squared end fraction straight R squared space minus 1 half straight u squared over straight g space equals space 0

rightwards double arrow space space space straight h space plus 1 half straight u squared over straight g space equals space fraction numerator straight g over denominator 2 straight u squared end fraction straight R squared space

rightwards double arrow space space space straight h space plus 1 half fraction numerator 2 gh over denominator straight g end fraction space equals space fraction numerator straight g over denominator 2 cross times 2 gh end fraction straight R squared space

rightwards double arrow space space straight R squared space equals 8 straight h

rightwards double arrow space space straight R space equals 2 square root of 2 space space straight h