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Sagot :
SOLUTION
Step 1 :
In this question, we have that :
[tex]h(t)=-16t^2\text{ + 140 t + 4 }[/tex]a) The first part of the question says:
when will the object reach the ground?
We need to set h ( t ) = 0 and then solve for t .
[tex]h(t)=-16t^2\text{ + 140 t + 4 = 0}[/tex][tex]\begin{gathered} We\text{ have that: } \\ 16t^2\text{ -140 t - 4 = 0 } \\ \text{Solving the quadratic equation, we have that:} \\ t_1=8.778seconds_{} \\ or_{} \\ t_2\text{ = -0.0028 seconds ( we ignore because it is negative)} \end{gathered}[/tex]The object will reach the ground in 8. 78 seconds ( 2 decimal places)
b ) When will the object reach its maximum height?
Recall that:
[tex]\begin{gathered} h(t)=-16t^2\text{ + 140 t + 4} \\ \text{Differentiating the function with respect to t, we have that:} \\ V\text{ ( t ) = }\frac{d\text{ h ( t )}}{d\text{ t }}\text{ = - 32 t + 140} \\ 0\text{ = - 32t + 140} \\ 32\text{ t = 140} \\ \text{Divide both sides by 32, we have that:} \\ \text{t = }\frac{140}{32} \\ \text{t = 4. 375 seconds} \end{gathered}[/tex]The object will reach its maximum height in 4. 38 seconds
( 2 decimal places)
c) What is its maximum height ?
[tex]\begin{gathered} h(t)=-16t^2\text{ + 140 t + 4} \\ \text{put t = 4.375 in h ( t ) , we have that:} \\ h(4.375)=-16(4.375)^2+140(4.375)^{}\text{ + 4} \\ h\text{ ( 4.375) = 310. 25 fe}et \end{gathered}[/tex]The maximum height is 310. 25 feet ( 2 decimal places)
d ) For how long will the object be above 206 feet?
Recall that:
[tex]\begin{gathered} h(t)=-16t^2\text{ + 140 t + 4} \\ \text{when h ( t ) = 206, we have that: 206 = -16t}^2\text{ + 140 t + 4} \\ 16t^2\text{ - 140 t + 206 -4 = 0} \\ 16t^2\text{ - 140t + 202 = 0} \end{gathered}[/tex]Solving the quadratic equation, we have that:
[tex]\begin{gathered} t_{\text{ 1}}\text{ = 6.928 seconds ( when the object is descending)} \\ t_2\text{ = 1.8224 seconds }(\text{ when the object is ascending)} \end{gathered}[/tex]The object be above 206 feet when the time, t is
[tex]\begin{gathered} (\text{ 4. 375 - 1.8224 ) x 2 } \\ =\text{ 2. 5526 x 2} \\ =\text{ 5. 1052 seconds} \\ =\text{ 5. 1 seconds ( 2 decimal places )} \end{gathered}[/tex]
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