The vertical component of the initial velocity is [tex]v_0_y = \frac{y}{t} + \frac{1}{2} gt[/tex]
The horizontal component of the initial velocity is [tex]v_0_x = \frac{x}{t}[/tex]
The horizontal displacement when the object reaches maximum height is [tex]X = \frac{xy}{gt^2} + \frac{x}{2}[/tex]
The given parameters;
the horizontal displacement of the object, = x
the vertical displacement of the object, = y
acceleration due to gravity, = g
time of motion, = t
The vertical component of the initial velocity is given as;
[tex]y = v_0_yt - \frac{1}{2} gt^2\\\\v_0_yt = y + \frac{1}{2} gt^2\\\\v_0_y = \frac{y}{t} + \frac{1}{2} gt[/tex]
The horizontal component of the initial velocity is calculated as;
[tex]x = v_0_xt\\\\v_0_x = \frac{x}{t}[/tex]
The time to reach to the maximum height is calculated as;
[tex]T = \frac{v_f_y -v_0_y}{-g} \\\\T = \frac{-v_0_y}{-g} \\\\T = \frac{v_0_y}{g} \\\\T = \frac{1}{g} (v_0_y)\\\\T = \frac{1}{g} (\frac{y}{t} + \frac{1}{2} gt)\\\\T = \frac{y}{gt} + \frac{1}{2} t[/tex]
The horizontal displacement when the object reaches maximum height is calculated as;
[tex]X= v_0_x \times T\\\\X= \frac{x}{t} \times (\frac{y}{gt} + \frac{1}{2} t)\\\\X = \frac{xy}{gt^2} + \frac{x}{2}[/tex]
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