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At a time t after being projected from the ground, a projectile is displaced x units horizontally and y units vertically above its point of projection. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration due to gravity.)

What are the horizontal and vertical components of the initial velocity of the projectile?

At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

Answer :

ONYEBUCHINNAJIGO

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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