Question

While standing on a 3-foot ladder, a grapefruit is tossed straight up with an initial velocity of 45 fu/sec. The initial position of the grapefruit is 7 feet above the ground when it is released.Its height at time t is given by y = h(t) = -16t^2 + 45t + 7.A) How high does it go before returning to the ground? Round time to 2 decimal places to compute height. ___feet.b) How long does it take the grapefruit to hit the ground? Round time to 3 decimal places. ___ seconds

Answer 1

ANSWER:

A) 38.64 feet.

B) 2.960 seconds

STEP-BY-STEP EXPLANATION:

We have that the function that models the situation of the statement is the following:

$h(t)=-16t^2+45t+7$

A)

We can calculate this height, which would be the maximum height it can reach before it begins to decay, as follows:

$\begin{gathered} t_v=-\frac{b}{2a} \\ \\ \text{ in this case b = 45, a = -16, we replacing:} \\ \\ t_v=-\frac{45}{-16\cdot2}=\frac{45}{32} \\ \\ t_v=1.406\text{ sec} \\ \\ \text{ Now, we replace in the function this time like this: } \\ \\ h(t)=-16\left(1.406\right)^2+45\left(1.406\right)+7 \\ \\ h(t)=38.64\text{ ft} \end{gathered}$

Therefore, the height is equal to 38.64 feet.

B)

To determine the time when the ground, we must make the height 0 and solve for t, just like this:

$\begin{gathered} 0=-16t^2+45t+7 \\ \\ -16t^2+45t+7=0 \\ \\ \text{ We use the general formula for quadratic equations} \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ a=-16,b=45,c=7 \\ \\ \text{ We replacing} \\ \\ t=\frac{-45\pm\sqrt{45^2-4(-16)(7)}}{2(-16)} \\ \\ t=\frac{-45\pm\sqrt{2473}}{-32} \\ \\ t_1=\frac{-45+\sqrt{2473}}{-32}=-0.148 \\ \\ t_2=\frac{-45-\sqrt{2473}}{-32}=\:2.960 \end{gathered}$

Therefore, the time is equal to 2.960 seconds