Question

Derive an expression for maximum height attained, time of flight, horizontal range for a projectile in oblique projection?

Answer 1

Maximum height (h_max): 

The maximum vertical distance travelled by the projectile during the journey is called maximum height. 

This is determined as follows: 

For the vertical part of the motion

\(u^2_y\) = \(u^2_y\) + 2a_ys

Here u_y = usinθ, a = -g, s = h_max, and at the maximum height v = 0

Time of flight (T_f): 

The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight. This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.

We know that S_y = u_y + \(\frac{1}{2}\)a_yt²

Here, s_y = y = 0 (net displacement in y-direction is zero), 

u_y = u sin θ, a = -g, t = T_f , Then

0 = usin θ T_f - \(\frac{1}{2}\)g\(T^2_f\)

T_f = 2u \(\frac {sin θ}{g}\) .....(2)

Horizontal range (R): 

The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write.

Range R = Horizontal component of velocity % time of flight = u cos θ × T_f = \(\frac{u^2sin2θ}{g}\)

The horizontal range directly depends on the initial speed (u) and the sine of angle of projection (θ). It inversely depends on acceleration due to gravity ‘g’.

For a given initial speed u, the maximum possible range is reached when sin 2θ is maximum, sin 2θ = 1. This implies 2θ = π/2 or θ = π/4

This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range, given by

R_max = \(\frac{u^2}{g}\).