JEE MAIN - Physics (2019 - 12th April Morning Slot - No. 1)
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) :
$${\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms-1
$${\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms-1
$${\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {3 \over 5}$$ ms-1
$${\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)$$ and $${v_0} = {5 \over 3}$$ ms-1
Explanation
Equation of trajectory is given as
y = 2x – 9x2 …(A)
Comparing with equation:
$$y = x\tan \theta - {g \over {2{u^2}{{\cos }^2}\theta }}{x^2}$$ ...(B)
We get, $$\tan \theta = 2$$
$$ \therefore \cos \theta = {1 \over {\sqrt 5 }}$$
Also, $${g \over {2{u^2}{{\cos }^2}\theta }} = 9$$
$$ \Rightarrow {{10} \over {2 \times 9 \times {{\left( {{1 \over {\sqrt 5 }}} \right)}^2}}} = {u^2};\,\,{u^2} = {{25} \over 9}$$
$$ \Rightarrow u = {5 \over 3}m/s$$
y = 2x – 9x2 …(A)
Comparing with equation:
$$y = x\tan \theta - {g \over {2{u^2}{{\cos }^2}\theta }}{x^2}$$ ...(B)
We get, $$\tan \theta = 2$$
$$ \therefore \cos \theta = {1 \over {\sqrt 5 }}$$
Also, $${g \over {2{u^2}{{\cos }^2}\theta }} = 9$$
$$ \Rightarrow {{10} \over {2 \times 9 \times {{\left( {{1 \over {\sqrt 5 }}} \right)}^2}}} = {u^2};\,\,{u^2} = {{25} \over 9}$$
$$ \Rightarrow u = {5 \over 3}m/s$$
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