Question 182: Motion in 2D & Vectors

Question

A projectile is fired at an angle of 45o with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

Accepted Answer

Let $$\phi $$ be elevation angle of the projectile at its
highest point as seen from the point of projection
O and $$	heta $$ be angle of projection with the horizontal.
From figure, tan$$\phi $$ = $${H \over {R/2}}$$      ....(1)
In case of projectile motion
Maximum height H = $${{{u^2}{{\sin }^2}	heta } \over {2g}}$$
Horizontal range, R = $${{{u^2}\sin 	heta } \over g}$$
Substituting these values of H and R in (1), we get
$$	an \phi  = {{{{{u^2}{{\sin }^2}	heta } \over {2g}}} \over {{{{u^2}\sin 2	heta } \over {2g}}}}$$
$$	an \phi  = {{{{\sin }^2}	heta } \over {\sin 2	heta }} = {{{{\sin }^2}	heta } \over {2\sin 	heta \cos 	heta }}$$
= $${1 \over 2}	an 	heta $$
$$	an \phi  = {1 \over 2}	an 45^\circ  = {1 \over 2}$$
Here, $$	heta $$ = 45ÃÂ°
$$ 	herefore $$ $$	an \phi  = {1 \over 2}	an 45^\circ  = {1 \over 2}$$     $$  \left(\because {	an 45^\circ  = 1} ight)$$
$$\phi  = {	an ^{ - 1}}\left( {{1 \over 2}} ight)$$

Answer

45o

Answer

60o

Answer

tan$$-$$1 $$\left( {{1 \over 2}} \right)$$

Answer

tan$$-$$1 $$\left( {{{\sqrt 3 } \over 2}} \right)$$

Motion in 2D & Vectors

A projectile is fired at an angle of 45o with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

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