Respuesta :
Question:
Britney throws an object straight up into the air with an initial velocity of 27 ft/s from a platform that is 10 ft above the ground. Use the formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height. How long will it take for the object to hit the ground?
1s 2s 3s 4s
Answer:
It takes 2 seconds for object to hit the ground
Solution:
The given equation is:
[tex]h(t) = -16t^2 + v_0t+h_0[/tex]
Initial velocity = 27 feet/sec
[tex]h_0 = 10\ feet[/tex]
Therefore,
[tex]h(t) = -16t^2 +27t+10[/tex]
At the point the object hits the ground, h(t) = 0
[tex]-16t^2 +27t+10 = 0\\\\16t^2-27t-10=0[/tex]
Solve by quadratic formula,
[tex]\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=16,\:b=-27,\:c=-10:\quad \\\\t=\frac{-\left(-27\right)\pm \sqrt{\left(-27\right)^2-4\cdot \:16\left(-10\right)}}{2\cdot \:16}\\\\t = \frac{27 \pm \sqrt{1369}}{32}\\\\t = \frac{27 \pm 37}{32}\\\\We\ have\ two\ solutions\\\\t = \frac{27+37}{32}\\\\t = \frac{64}{32}\\\\t = 2\\\\And\\\\t = \frac{27-37}{32}\\\\t = -0.3125[/tex]
Ignore, negative value
Thus, it takes 2 seconds for object to hit the ground
It will take 2 seconds for the object to hit the ground.
Given that,
Britney throws an object straight up into the air with an initial velocity of 27 feet/sec.
From a platform that is 10 feet above the ground.
Formula; [tex]h(t)= -16t^2+v_ot+h_0[/tex],
Where [tex]v_0[/tex] is the initial velocity and [tex]h_0[/tex] is the initial height.
We have to determine,
How long will it take for the object to hit the ground.
According to the question,
Formula; [tex]h(t)= -16t^2+v_ot+h_0[/tex]
Where,
[tex]h_0 = 10 \ feet \ and \\ v_0 = 27\ feet \ per \ second[/tex]
Substitute the values in the given formula,
[tex]h(t)= -16t^2+v_ot+h_0\\\\h(t) = -16t^2 + 27t + 10[/tex]
When the point the object hits the ground, h(t) become 0,
Then,
[tex]\h(t) = -16t^2 + 27t + 10\\\\0 = -16t^2 + 27t + 10\\\\ -16t^2 + 27t + 10 = 0[/tex]
For the quadratic equation solution is determined by the formula.
[tex]x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\[/tex]
Where, a = -16, b = 27, c = 10
Substitute all the values in the formula;
[tex]x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\\x = \dfrac{-(-27)\pm\sqrt{(-27)^2-4(16).(10)}}{2(16)}\\\\\\x = \dfrac{27\pm\sqrt{729+640}}{32}\\\\x = \dfrac{27\pm\sqrt{1369}}{32}}\\\\x = \dfarc{27\pm37}{32}\\\\x = \dfrac{27+37}{32} \ and \ x = \dfrac{27-37}{32}\\\\x = \dfrac{64}{32} \ and \ x= \dfrac{-10}{32}\\\\x = 2 \ and \ x = - 0.3125[/tex]
Time can not be negative. so, x = -0.3125 rejected.
Hence, It will take 2 seconds for the object to hit the ground.
To know more about Quadratic equation click the link given below.
https://brainly.com/question/14870558
