Question

Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity at an angel θ to the horizontal.

Answer 1

1. Consider a body projected with velocity \(\vec{u},\) at an angle θ of projection from point O in the co-ordinate system of the XY- plane, as shown in figure.

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

u_x = u cos θ (Horizontal component) 

u_y = u sin θ (Vertical component)

3. 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. Thus, 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 = usin θ – gt

5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,

6. At the highest point, the time of ascent of the projectile is given as,

7. The total time in air i.e., time of flight is given as,

8. The total horizontal distance travelled by the particle in this time T is given as,

This is required expression for horizontal range of the projectile.

Expression for maximum height of a projectile: The maximum height H reached by the projectile is the distance travelled along the vertical (y) direction in time t_A.

Substituting s_y = H and t = t_a in equation (1), we have,

This equation represents maximum height of projectile.