Question

A particle is projected from a point A at an angle(Q) with the horizontal. At B it moves at right angle to its initial direction. Find time of Flight from A to B.

Answer 1

A particle is projected from a point A at an angle(Q) with the horizontal. At B it moves at right angle to its initial direction. Find time of Flight from A to B.

$\overrightarrow{v_0}$ - vector of initial velocity

$\vec{v}$ - vector of current velocity

Suppose, at the time instant $t \overrightarrow{v_0}$ is perpendicular to $\vec{v}$ . Then:

$$\overrightarrow {v _ {0}} * \vec {v} = 0$$

On the other hand:

$$\vec {v} = \overrightarrow {v _ {0}} + \vec {g} t,$$

where $\vec{g}$ - gravitational acceleration.

Therefore:

$$(\overrightarrow {v _ {0}} + \vec {g} t) * \overrightarrow {v _ {0}} = 0$$ $$\overrightarrow {v _ {0}} ^ {2} + (\vec {g} * \overrightarrow {v _ {0}}) t = v _ {0} ^ {2} + g v _ {0} \cos (a) t = 0$$

where $a$ - angle between $\vec{g}$ and $\overrightarrow{v_0}$ , $a = 90 + Q$

$$v _ {0} ^ {2} - g v _ {0} \sin (Q) t = 0$$ $$t = \frac {v _ {0}}{g \sin (Q)}$$

Answer: $t = \frac{v_0}{g \sin(Q)}$