Question

The direction of motion of a projectile at a certain instant is inclined at a angle 60 to the horizontal. After root 3 second it is inclined an angle 30. The horizontal component of velocity of projectile is?

Answer 1

The direction of motion of a projectile at a certain instant is inclined at a angle 60 to the horizontal. After root 3 second it is inclined an angle 30. The horizontal component of velocity of projectile is?

Solution

$$\alpha_ {1} = 60, \alpha_ {2} = 30, t = \sqrt {3},$$ $$\frac {v _ {y}}{v _ {x}} = \tan \alpha_ {1}, \frac {v _ {y ^ {1}}}{v _ {x}} = \tan \alpha_ {2}, v _ {y ^ {1}} = v _ {y} - g t$$

Then

$$v _ {y ^ {1}} = v _ {x} \tan \alpha_ {2} = v _ {x} \tan \alpha_ {1} - g t$$ $$v _ {x} = \frac {g t}{\tan \alpha_ {1} - \tan \alpha_ {2}} = \frac {10 \sqrt {3}}{\sqrt {3} - \frac {\sqrt {3}}{3}} = 15 \frac {m}{s}$$