Source: Principles of Physics
Problem
A projectile is launched from ground level. At a time $t_1$, its horizontal displacement is $x_1$ and its vertical displacement is $y_1$.
- What is the horizontal component of the initial velocity, $v_{0x}$?
- What is the vertical component of the initial velocity, $v_{0y}$?
- What is its horizontal displacement, $x_{peak}$, when it reaches its maximum height?
[Q1] $v_{0x} = \frac{x_1}{t_1}$ [Q2] $v_{0y} = \frac{y_1}{t_1} + \frac{1}{2}gt_1$ [Q3] $x_{peak} = \frac{x_1 y_1}{gt_1^2} + \frac{x_1}{2}$
The kinematic equations for projectile motion are $x(t) = v_{0x} t$ and $y(t) = v_{0y} t - \frac{1}{2}gt^2$.
From the horizontal motion equation at $t=t_1$:
$$x_1 = v_{0x} t_1 \implies v_{0x} = \frac{x_1}{t_1}$$From the vertical motion equation at $t=t_1$:
$$y_1 = v_{0y} t_1 - \frac{1}{2}gt_1^2 \implies v_{0y} = \frac{y_1}{t_1} + \frac{1}{2}gt_1$$Maximum height is reached when the vertical velocity $v_y = v_{0y} - gt$ becomes zero. The time to reach this peak is $t_{peak} = v_{0y}/g$.
The horizontal displacement at this time is $x_{peak} = v_{0x} t_{peak}$. Substituting the derived expressions for $v_{0x}$ and $v_{0y}$:
$$x_{peak} = v_{0x} \left( \frac{v_{0y}}{g} \right) = \left( \frac{x_1}{t_1} \right) \frac{1}{g} \left( \frac{y_1}{t_1} + \frac{1}{2}gt_1 \right) = \frac{x_1 y_1}{gt_1^2} + \frac{x_1}{2}$$