Question

an object is thrown upward at a speed of 98 feet per second by a machine from a height of 3 feet off the ground. the height h of the object after t seconds can be found using the equation h = - 16t² + 98t + 3. when will the height be 53 feet? select an answer. when will the object reach the ground? select an answer. round all answers to one decimal place. question help: post to forum. submit question jump to answer

Explanation:

Step1: Set up equation for height 53 feet

Set $h = 53$ in $h=-16t^{2}+98t + 3$, so $-16t^{2}+98t + 3=53$. Rearrange to get $-16t^{2}+98t - 50 = 0$. Divide by -2: $8t^{2}-49t + 25=0$.

Step2: Use quadratic formula

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 8$, $b=-49$, $c = 25$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute values: $t=\frac{49\pm\sqrt{(-49)^{2}-4\times8\times25}}{2\times8}=\frac{49\pm\sqrt{2401 - 800}}{16}=\frac{49\pm\sqrt{1601}}{16}$. Calculate $\sqrt{1601}\approx40.01$. Then $t=\frac{49\pm40.01}{16}$. We get $t_1=\frac{49 + 40.01}{16}\approx5.6$ and $t_2=\frac{49 - 40.01}{16}\approx0.6$.

Step3: Set up equation for reaching ground

Set $h = 0$ in $h=-16t^{2}+98t + 3$, so $-16t^{2}+98t + 3=0$. Here $a=-16$, $b = 98$, $c = 3$.

Step4: Apply quadratic formula again

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-98\pm\sqrt{98^{2}-4\times(-16)\times3}}{2\times(-16)}=\frac{-98\pm\sqrt{9604+192}}{-32}=\frac{-98\pm\sqrt{9796}}{-32}$. Calculate $\sqrt{9796}\approx98.97$. Then $t=\frac{-98\pm98.97}{-32}$. We have $t_1=\frac{-98 + 98.97}{-32}\approx - 0.03$ (rejected as time can't be negative) and $t_2=\frac{-98-98.97}{-32}\approx6.2$.

Answer:

When the height is 53 feet, $t\approx0.6$ s and $t\approx5.6$ s. When the object reaches the ground, $t\approx6.2$ s.