Question

For problem below, make use of the equation of motion of a vertically projected object: 20012 gtv t s  , where sis the height of the object at time tseconds, gis 32 feet per second² and 0s and 0s are the initial velocity and initial height, respectively.Two friends are in competition. Each stands on the roof of the apartment building where each one lives. At the same time, each friend stands at the edge of the roof and throws up a basketball as hard as possible. Joan’s apartment building is 120 feet high, and she throws the basketball at a speed of 20feet per second. Joe’s apartment building is 105 feet high, and he throws the basketball at a speed of 34feet per second.a)Write the function that describes the height of Joan’s basketball, in terms of time measured in seconds. What kind of function is the height function

Answer 1

Answer:

$s=-4.905*t^{2} +20*t+120$

The equation corresponds to a quadratic function.

Step-by-step explanation:

Given

$s=-\frac{1}{2} gt^{2} +v_{0} t+s_{0}$

where

$g=9.81\frac{m}{s^{2} }\\ v_{0}=20\frac{ft}{s} \\ s_{0} =120 ft$

the function that describes the height of Joan’s basketball is

$s=-\frac{1}{2} (9.81\frac{m}{s^{2} })t^{2} +20\frac{ft}{s}*t+120 ft\\ s=-4.905*t^{2} +20*t+120$

The equation corresponds to a quadratic function.