JEE MAIN - Physics (2012 - No. 22)
A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy
can throw the same stone up to will be
$$20\sqrt 2 $$ m
10 m
$$10\sqrt 2 $$ m
20 m
Explanation
We know, $$R = {{{u^2}{{\sin }2}\theta } \over g}$$ and $$H = {{{u^2}{{\sin }^2}\theta } \over {2g}};$$
$${H_{\max }}\,\,$$ is possible when $$\theta = 90$$$$^\circ $$
$${H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2$$
As $$R = {{{u^2}\sin 2\theta } \over g}$$
Range is maximum when projectile is thrown at an angle $$45^\circ $$.
$$ \Rightarrow {R_{\max }} = {{{u^2}} \over g}$$
$${R_{\max }} = {{10 \times g \times 2} \over g} = 20$$ meter
$${H_{\max }}\,\,$$ is possible when $$\theta = 90$$$$^\circ $$
$${H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2$$
As $$R = {{{u^2}\sin 2\theta } \over g}$$
Range is maximum when projectile is thrown at an angle $$45^\circ $$.
$$ \Rightarrow {R_{\max }} = {{{u^2}} \over g}$$
$${R_{\max }} = {{10 \times g \times 2} \over g} = 20$$ meter
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