Question 107: Motion in 2D & Vectors

Question

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2
. If it were launched at an
angle $$	heta $$0 with speed v0 then (g = 10 ms–2) :

Accepted Answer

Equation of trajectory is given as
y = 2x Ã¢ÂÂ 9x
2
Ã¢ÂÂ¦(A)
Comparing with equation:
$$y = x	an 	heta  - {g \over {2{u^2}{{\cos }^2}	heta }}{x^2}$$    ...(B)
We get, $$	an 	heta  = 2$$
$$ 	herefore \cos 	heta  = {1 \over {\sqrt 5 }}$$
Also, $${g \over {2{u^2}{{\cos }^2}	heta }} = 9$$
$$ \Rightarrow {{10} \over {2 	imes 9 	imes {{\left( {{1 \over {\sqrt 5 }}} ight)}^2}}} = {u^2};\,\,{u^2} = {{25} \over 9}$$
$$ \Rightarrow u = {5 \over 3}m/s$$

Answer

$${	heta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} ight)$$ and $${v_0} = {5 \over 3}$$ ms-1

Answer

$${	heta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} ight)$$ and $${v_0} = {3 \over 5}$$ ms-1

Answer

$${	heta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} ight)$$ and $${v_0} = {3 \over 5}$$ ms-1

Answer

$${	heta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} ight)$$ and $${v_0} = {5 \over 3}$$ ms-1

Motion in 2D & Vectors

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2 . If it were launched at an angle $\theta $0 with speed v0 then (g = 10 ms–2) :

Choose the correct answer: