Answer:
a) The path is usually modeled as a parabola. (In real life, it is most definitely not a parabola, as air resistance and "flight" are definite factors.)
b) To get maximum height for a given speed, the ball should be launched straight up. (In real life, the kicker's effective speed will vary with launch angle.)
Step-by-step explanation:
The equations modeling air resistance and/or the fluid dynmics of an odd shape through air are impossible to solve "by hand", so we use equations that we can solve. For many purposes (teaching equations, functions, and solution processes, for example), the equation of a parabola is "good enough."
Vertical height is determined by vertical speed. For a given launch angle, the vertical speed is the launch speed multiplied by the sine of the angle (relative to the ground). The sine is a maximum when the angle is 90°, or straight up.
Answer:
The path of the ball that has been punted is parabola.
Maximum height is reached when angle is 90°.
Step-by-step explanation:
The path of the ball that has been punted is parabola.
This is an example of projectile motion.
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g [tex]m/s^2[/tex] and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 [tex]m/s^2[/tex] and time taken = 2u sin θ /g
So range of projectile, [tex]R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}[/tex]
Vertical motion (Maximum height reached, H) :
We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
[tex]0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}[/tex]
So value of maximum height is proportional to sin²θ.
sin²θ is maximum when θ = 90°
So maximum height is reached when θ = 90° that is straight upward.
Maximum height is reached when angle is 90°.
