Question 69: Motion in 2D & Vectors

Question

Two projectiles P1 and P2 thrown with speed in the ratio $$\sqrt3$$ : $$\sqrt2$$, attain the same height during their motion. If P2 is thrown at an angle of 60$$^\circ$$ with the horizontal, the angle of projection of P1 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 P1 and P2 thrown with speed in the ratio $\sqrt3 : \sqrt2, attain the same height during their motion. If P2 is thrown at an angle of 60^\circ$ with the horizontal, the angle of projection of P1 with horizontal will be :

Choose the correct answer: