The position of a projectile launched from the origin at t = 0 is given by r → = 4 0  i ⁡ ^ + 5 0  j ⁡ ^ m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms -2 ).

Question

The position of a projectile launched from the origin at t = 0 is given by r &#8594; = 4 0&#160; i &#8289; ^ + 5 0&#160; j &#8289; ^ m at t = 2s. If the projectile was launched at an angle &#952; from the horizontal, then &#952; is (take g = 10 ms -2 ).

Aicharya · Accepted Answer

From question, Horizontal velocity (initial),
$u_{x} = \frac{40}{2} = 20 m / s$
Vertical velocity (initial), $50 = u_{y} t + \frac{1}{2} g t^{2}$

$\Rightarrow 50 = u_{y} \times 2 + \frac{1}{2} (- 10) \times 4$
$50 = 2 u_{y} - 20$
$u_{y} = \frac{70}{2} = 35 m / s$
$∴ \tan θ = \frac{u_{y}}{u_{x}} = \frac{35}{20} = \frac{7}{4}$
$\Rightarrow$ $θ = \tan^{- 1} \frac{7}{4}$

The position of a projectile launched from the origin at t = 0 is given by r → = 4 0  i ⁡ ^…

Solution

The position of a projectile launched from the origin at t = 0 is given by r &#8594; = 4 0&#160; i &#8289; ^…

Solution

The position of a projectile launched from the origin at t = 0 is given by r → = 4 0 i ⁡ ^…