A 0.45-kg soccer ball is kicked at an angle θ above the ground with initial speed vi. Determine the height of the ball as a function of the horizontal distance d it travels down the field.
d*tan(θ) - ((gd^(2))/(2vi^(2)(cos(θ))^(2))
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To determine the height of the ball as a function of the horizontal distance it travels down the field, we can use the concept of projectile motion. In projectile motion, the motion of the object can be divided into horizontal and vertical components.
Let's break down the problem into its components:
1. Horizontal Motion:
The horizontal component of the velocity remains constant throughout the ball's flight. Therefore, the horizontal distance traveled can be calculated using the equation:
d = vi * t,
where d is the horizontal distance, vi is the initial horizontal velocity, and t is the time of flight.
2. Vertical Motion:
The vertical motion of the ball can be investigated using a free fall motion because the only force acting in this direction is the gravitational force. We can use the kinematic equations to relate the vertical distance, time, initial vertical velocity, and acceleration due to gravity.
The vertical distance traveled can be calculated using the equation:
h = vi * sin(θ) * t - (1/2) * g * t^2,
where h is the height, vi is the initial vertical velocity, θ is the angle above ground, t is the time of flight, and g is the acceleration due to gravity.
To find the time of flight, we can use the equation for vertical displacement and initial vertical velocity:
0 = vi * sin(θ) * t - (1/2) * g * t^2.
By rearranging this equation, we get:
t = (vi * sin(θ)) / g.
Now, substitute the value of t into the equation for the horizontal distance:
d = vi * [(vi * sin(θ)) / g].
Consequently, the height of the ball as a function of the horizontal distance traveled down the field (h(d)) is:
h(d) = vi * sin(θ) * [(vi * sin(θ)) / g] - (1/2) * g * [(vi * sin(θ)) / g]^2.
Simplifying further will yield the final expression for h(d).