Question 80: Motion in 2D & Vectors

Question

Two projectiles P
1
and P
2
thrown with speed in the ratio $\sqrt3$ : $\sqrt2$, attain the same height during their motion. If P
2
is thrown at an angle of 60$^\circ$ with the horizontal, the angle of projection of P
1
with horizontal will be :

Accepted Answer

We know,
Maximum height of a projectile $(H) = {{{u^2}{{\sin }^2}	heta } \over {2g}}$
Given, Ratio of initial velocity of two projectile
${{{u_1}} \over {{u_2}}} = {{\sqrt 3 } \over {\sqrt 2 }}$
Both projectile reach the same maximum height.
$	herefore$ ${H_1} = {H_2}$
$ \Rightarrow {{u_1^2{{\sin }^2}{	heta _1}} \over {2g}} = {{u_2^2{{\sin }^2}{	heta _2}} \over {2g}}$
$ \Rightarrow u_1^2{\sin ^2}{	heta _1} = u_2^2{\sin ^2}{	heta _2}$
$ \Rightarrow {\left( {{{{u_1}} \over {{u_2}}}} ight)^2} = {\left( {{{\sin {	heta _2}} \over {\sin {	heta _1}}}} ight)^2}$
$ \Rightarrow {\left( {{{\sqrt 3 } \over {\sqrt 2 }}} ight)^2} = {\left( {{{\sin 60^\circ } \over {\sin {	heta _1}}}} ight)^2}$
$ \Rightarrow {3 \over 2} = {3 \over {4{{\sin }^2}{	heta _1}}}$
$ \Rightarrow \sin _{{	heta _1}}^2 = {1 \over 2}$
$ \Rightarrow \sin {	heta _1} = {1 \over {\sqrt 2 }} = \sin 45^\circ $
$ \Rightarrow {	heta _1} = 45^\circ $

Answer

15$^\circ$

Answer

30$^\circ$

Answer

45$^\circ$

Answer

60$^\circ$

Motion in 2D & Vectors

Two projectiles P 1 and P 2 thrown with speed in the ratio \sqrt3 : \sqrt2, attain the same height during their motion. If P 2 is thrown at an angle of 60^\circ with the horizontal, the angle of projection of P 1 with horizontal will be :

Choose the correct answer: