Answer:
a) The path is usually modeled as a parabola. (In real life, it is most definitely <em>not</em> 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 <em>can</em> 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.
6s+6t is the answer that won't let me write
Answer:
THIS KID DON'T MAKE NO SICES
Step-by-step explanation:
ANSWER
C) 5.7 seconds
EXPLANATION
The height of the object is given by:
![h(t) = - 16 {t}^{2} + 64t + 160](https://tex.z-dn.net/?f=h%28t%29%20%3D%20%20-%2016%20%7Bt%7D%5E%7B2%7D%20%20%2B%2064t%20%2B%20160)
If the object hit the ground, then the height is zero.
![- 16 {t}^{2} + 64t + 160 = 0](https://tex.z-dn.net/?f=-%2016%20%7Bt%7D%5E%7B2%7D%20%20%2B%2064t%20%2B%20160%20%3D%200)
Divide through by -16
![{t}^{2} - 4t - 10 = 0](https://tex.z-dn.net/?f=%20%7Bt%7D%5E%7B2%7D%20%20%20-%204t%20%20-%2010%20%3D%200)
Where a=1, b=-4 and c=-10
We substitute into the quadratic formula to obtain,
![t= \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}](https://tex.z-dn.net/?f=t%3D%20%20%5Cfrac%7B%20-%20b%20%5Cpm%20%5Csqrt%7B%20%7Bb%7D%5E%7B2%7D%20%20-%204ac%7D%20%7D%7B2a%7D%20)
![t= \frac{ - - 4\pm \sqrt{ {( - 4)}^{2} - 4( 1)( - 10)} }{2(1)}](https://tex.z-dn.net/?f=t%3D%20%20%5Cfrac%7B%20-%20%20-%204%5Cpm%20%5Csqrt%7B%20%7B%28%20-%204%29%7D%5E%7B2%7D%20%20-%204%28%201%29%28%20-%2010%29%7D%20%7D%7B2%281%29%7D%20)
![t= \frac{ 4\pm \sqrt{56} }{2}](https://tex.z-dn.net/?f=t%3D%20%20%5Cfrac%7B%204%5Cpm%20%5Csqrt%7B56%7D%20%7D%7B2%7D%20)
![t= \frac{ 4\pm 2\sqrt{14} }{2}](https://tex.z-dn.net/?f=t%3D%20%20%5Cfrac%7B%204%5Cpm%202%5Csqrt%7B14%7D%20%7D%7B2%7D%20)
t=2-√14 or t=2+√14
Time cannot be negative.
Hence, t=5.7 seconds to the nearest tenth.