Question 104: Motion in 2D & Vectors

Question

The trajectory of a projectile in a vertical plane is y = $\alpha$x $-$ $\beta$x
2
, 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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