A projectile is launched at an angle α with the horizontal with a velocity 20   m   s - 1 . After 10   s , its inclination with horizontal is β . The value of tan β will be : g = 10   m   s - 2 .

Question

A projectile is launched at an angle &#945; with the horizontal with a velocity 20 &#160; m &#160; s - 1 . After 10 &#160; s , its inclination with horizontal is &#946; . The value of tan &#946; will be : g = 10 &#160; m &#160; s - 2 .

Aicharya · Accepted Answer

Component of velocity in horizontal direction,

$v_{x} = 20 \cos α = v_{2} \cos β$

Component of velocity in vertical direction,

$v_{y} = 20 \sin α - g \times 10 = v_{2} \sin β$

$\Rightarrow \tan β = \frac{20 \sin α - 100}{20 \cos α}$

$\Rightarrow \tan β = \tan α - \frac{5}{\cos α}$

$\Rightarrow \tan β = \tan α - 5 \sec α$

A projectile is launched at an angle α with the horizontal with a velocity 20   m   s - 1 .…

Solution

A projectile is launched at an angle &#945; with the horizontal with a velocity 20 &#160; m &#160; s - 1 .…

Solution

A projectile is launched at an angle α with the horizontal with a velocity 20 m s - 1 .…