Question

(a)a body is projected with speed u at an angle to the horizontal to have maximum range. What is the velocity at the highest point?(b) What is the angle of projection of a projectile motion whose range R is n times the maximum hight.

Answer 1

Answer:

A) Zero

B) $\tan ^{ -1 }{\left( \frac{ 4 }{ n }\right)}$

a) Let us consider the body is projected to the maximum height h = h_{max}  

When the body reaches the highest point, due to Newton's law of gravity, its velocity drops to zero and then it begins to fall to the ground. Hence at highest point, h_{max}, the velocity will be zero.  

b) maximum range of the body R is n times the h_{max}  

             $R=nh_{ max }\quad .....(1)$

We know that, $R=\frac { u^{ 2 }\sin { 2a } }{ g }$

             $h=\frac { u^{ 2 }sin^{ 2 }a }{ 2g }$

             $\frac { u^{ 2 }\sin { 2a } }{ g } =\frac { nu^{ 2 }sin^{ 2 }a }{ 2g }$

             $\sin { 2a } =\frac { n\sin ^{ 2 }{ a } }{ 2 }$

             $2\sin { a } \cos { a } =\frac { n\sin ^{ 2 }{ a } }{ 2 }$

             $\frac { 4 }{ n } =\frac { \sin { a}}{ \cos { a }}$

             $a=\tan ^{ -1 }{ \left( \frac { 4 }{ n} \right)}$

Angle of projection, $a=\tan ^{ -1 }{ \left(\frac { 4}{ n}\right)}$

Answer 2

Answer:

Explanation:

Hope it helps everyone