Question

Answer the following question.

Derive the formula for the range and maximum height achieved by a projectile thrown from the origin with initial velocity `vec"u"` at an angle 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 coordinate system of the XY- plane, as shown in the 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 sin θ
4. Thus, the components of velocity of the projectile at time t are given by,
   v_x = u_x = u cosθ
   v_y = u_y - gt = u sin θ - gt
5. Similarly, the components of displacements of the projectile in the horizontal and vertical directions at time t are given by,
   s_x = (u cos θ)t 
   s_y = (u sin θ)t `- 1/2 "gt"^2`      ....(1)
6. At the highest point, the time of ascent of the projectile is given as, t_A = `("u sin"theta)/"g"`   ....(2)
7. The total time in air i.e., time of flight is given as,
   T = 2t_A = `(2"u"sin theta)/"g"`    .....(3)
8. The total horizontal distance travelled by the particle in this time T is given as,
   R = `"u"_"x" * "T"`
   R = `u cos θ * (2t_A)`
   ∴ R = u cos θ `* ("2u" sin theta)/"g"`    ....[From(3)]
   ∴ R = `("u"^2(2 sin theta * cos theta))/"g"`
   ∴ R = `("u"^2 sin 2 theta)/"g"`   ....[∵ sin 2θ = 2 sin θ . cos θ]
   This is a required expression for the horizontal range of the projectile. 

Expression for a maximum height of a projectile:

1. The maximum height H reached by the projectile is the distance travelled along the vertical (y) direction in time t_A.
2. Substituting s_y = H and t = t_a in equation (1), we have,
   H = `("u"sin theta)"t"_"A" - 1/2"gt"_"A"^2`
   ∴ H = `"u" sin theta (("u sin"theta)/"g") - 1/2 "g"(("u" sin theta)/"g")^2`     ....[From (2)]
   ∴ H = `("u"^2sin^2theta)/"2g" = "u"_"y"^2/"2g"`
   This equation represents the maximum height of projectile.