Question 93: Motion in 2D & Vectors

Question

The trajectory of a projectile in a vertical plane is y = $$\alpha$$x $$-$$ $$\beta$$x2, where $$\alpha$$ and $$\beta$$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $$	heta$$ and the maximum height attained H are respectively given by :

Accepted Answer

y = $$\alpha$$x $$-$$ $$\beta$$x
2
comparing with trajectory equation
$$y = x	an 	heta  - {1 \over 2}{{g{x^2}} \over {{u^2}{{\cos }^2}	heta }}$$
$$	an 	heta  = \alpha  \Rightarrow 	heta  = {	an ^{ - 1}}\alpha $$
$$\beta  = {1 \over 2}{g \over {{u^2}{{\cos }^2}	heta }}$$
$$ \Rightarrow $$ $${u^2} = {g \over {2\beta {{\cos }^2}	heta }}$$
Maximum height H :
$$H = {{{u^2}{{\sin }^2}	heta } \over {2g}} = {g \over {2\beta {{\cos }^2}	heta }}{{{{\sin }^2}	heta } \over {2g}}$$
$$ \Rightarrow $$ $$H = {{{{	an }^2}	heta } \over {4\beta }} = {{{\alpha ^2}} \over {4\beta }}$$

Answer

$${	an ^{ - 1}}\alpha ,{{{\alpha ^2}} \over {4\beta }}$$

Answer

$${	an ^{ - 1}}\alpha ,{{4{\alpha ^2}} \over \beta }$$

Answer

$${	an ^{ - 1}}\left( {{\beta  \over \alpha }} ight),{{{\alpha ^2}} \over \beta }$$

Answer

$${	an ^{ - 1}}\beta ,{{{\alpha ^2}} \over {2\beta }}$$

Motion in 2D & Vectors

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