Respuesta :
We are given that the height of an object that is moving in projectile motion is the following:
[tex]S=-16t^2+v_0t[/tex]Where:
[tex]\begin{gathered} S=\text{ height} \\ t=\text{ time} \\ v_0=\text{ initial velocity} \end{gathered}[/tex]Now we will substitute the value of S for 272 ft, we get:
[tex]272=-16t^2+v_0t[/tex]Now we subtract 272 from both sides, we get:
[tex]-16t^2+v_0t-272=0[/tex]We get an equation of the form:
[tex]at^2+bt+c=0[/tex]The value of "t" are determined using the quadratic formula:
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Substituting the values we get:
[tex]t=\frac{-v_0\pm\sqrt[]{v^2_0-4(-16)(-272)}}{2(-16)}[/tex]Solving the operations we get:
[tex]t=\frac{-v_0\pm\sqrt[]{v^2_0-17408}}{-32}[/tex]Therefore, we get two possible values for the time:
[tex]t_1=\frac{-v_0+\sqrt[]{v^2_0-17408}}{-32}[/tex]The second value is:
[tex]t_2=\frac{-v_0-\sqrt[]{v^2_0-17408}}{-32}[/tex]Part b. We are asked to determine the time when the object returns to the ground and the initial velocity is 32 ft/s. To do that we will use the formula for the height:
[tex]S=-16t^2+v_0t[/tex]Now we substitute the value of the initial speed:
[tex]S=-16t^2+32t[/tex]Now, since we want the time when the object returns to the ground this means that the height must be zero, therefore, we substitute the values S = 0, we get:
[tex]0=-16t^2+32t[/tex]Now we take "t" as a common factor:
[tex]0=t(-16t+32)[/tex]Now we set each factor to zero:
[tex]t=0[/tex]For the second factor we get:
[tex]-16t+32=0[/tex]Now we subtract 32 from both sides:
[tex]-16t=-32[/tex]Now we divide both sides by -16:
[tex]t=-\frac{32}{-16}[/tex]Solving the operations:
[tex]t=2[/tex]Therefore, in 2 seconds the object will return to the ground.
