Question

A projectile is projected from point 'O' inside a vertical triangle ABC with base angles 30° and 60°. Projectile is projected in such a way that it touches at points P and Q of triangle as shown in figure. Take horizontal speed of projectile as $\sqrt{3}m/s$.

Answer 1

0.4 s

Answer 2

The problem describes a projectile motion where the trajectory touches two sides of a triangle, which act as tangents to the parabolic path. We are given the horizontal speed of the projectile and the angles of the tangents.

1. Identify the given information:

   • Horizontal speed of the projectile, $v_x = \sqrt{3} \text{ m/s}$. The horizontal component of velocity remains constant throughout the projectile motion.
   • The side AB is tangent to the trajectory at point P. The angle of AB with the horizontal base AC is $30^\circ$.
   • The side BC is tangent to the trajectory at point Q. The angle of BC with the horizontal base AC is $60^\circ$.

2. Determine the vertical velocity at point P: At point P, the velocity vector of the projectile is along the tangent AB. The angle of the velocity vector with the horizontal is $30^\circ$. The slope of the tangent is given by $\frac{v_y}{v_x}$. So, $\tan(30^\circ) = \frac{v_{yP}}{v_x}$. $v_{yP} = v_x \tan(30^\circ) = \sqrt{3} \text{ m/s} \times \frac{1}{\sqrt{3}} = 1 \text{ m/s}$.

3. Determine the vertical velocity at point Q: At point Q, the velocity vector of the projectile is along the tangent BC. Since the projectile is moving downwards at Q (after crossing the highest point), the angle of the velocity vector with the horizontal is $-60^\circ$. So, $\tan(-60^\circ) = \frac{v_{yQ}}{v_x}$. $v_{yQ} = v_x \tan(-60^\circ) = \sqrt{3} \text{ m/s} \times (-\sqrt{3}) = -3 \text{ m/s}$.

4. Calculate the time taken from P to Q: For vertical motion under gravity, the change in vertical velocity is given by $\Delta v_y = v_{y, \text{final}} - v_{y, \text{initial}} = -g \Delta t$. Here, $v_{y, \text{final}} = v_{yQ} = -3 \text{ m/s}$ and $v_{y, \text{initial}} = v_{yP} = 1 \text{ m/s}$. Let $T_{PQ}$ be the time taken from P to Q. $\Delta v_y = v_{yQ} - v_{yP} = -3 - 1 = -4 \text{ m/s}$. Therefore, $-4 = -g T_{PQ}$. $T_{PQ} = \frac{4}{g}$.

5. Substitute the value of g: Assuming the standard value of acceleration due to gravity, $g = 10 \text{ m/s}^2$. $T_{PQ} = \frac{4}{10} = 0.4 \text{ s}$.

The time taken by the projectile in moving from point P to Q is $0.4 \text{ s}$.