Question 188: Motion in 2D & Vectors

Question

A projectile is fired at an angle of 45
o
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

45
o

Answer

60
o

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

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