Question

Q.a) What is the trajectory followed by a projectile. b) What are the vertical and horizontal components of velocity? c) Derive an expression for time of flight.

Answer 1

(a) Parabolic

(b) Vertical component is vsinθ and Horizontal component is vcosθ

(c)

1. Consider a body projected with velocity \(\overset\rightarrow{u}\), at an angle θ of projection from point O in the coordinate system of the X-Y plane, as shown in figure.

2. The initial velocity \(\overset\rightarrow{u}\) can be resolved into two rectangular components:

u_x = ucos θ (Horizontal component)

u_y = usin θ (Vertical component)

3. Thus, the horizontal component remains constant throughout the motion due to the absence of any force acting in that direction, while the vertical component changes according to, v_y = u_y + a_yt with a_y = – g and u_y = u sinθ

 4. The components of velocity of the projectile at time t are given by, v_x = u_x = u cos θ

v_y = u_y – gt = usinθ – gt ... (1)

5. At maximum height, v_y = 0, t = t_A = time of ascent = time taken to reach maximum height.

∴ 0 = usinθ - gt_A ... (From 1)

∴ t_A = usinθ / g

6. Total time of flight:- 

T = 2t_A = \(\frac{2usinθ}{g} \)