Question

4: A tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 120 feet per second. What is the maximum height, in feet, the ball will attain?

Round to the nearest whole foot. Answer: _____

5: A tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 110 feet per second. After how many seconds does the ball attain its maximum height?

Round to the nearest hundredth. Answer: ____

6: The finishing time for a runner completing the 200-meter dash is affected by the tail-wind speed, s . The change, t, in a runner's performance is modeled by the function shown below:

t = 0.0119s^2 – 0.308s – 0.0003

Predict the change in a runner's finishing time with a wind speed of 5 meters/second. Note: A negative answer means the runner finishes with a lower time.

Round to the nearest hundredths. Answer: ____

Answer 1

Answer on Question #55823 – Math – Calculus

4. A tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 120 feet per second. What is the maximum height, in feet, the ball will attain? Round to the nearest whole foot.

Solution

$\mathrm{h}(t)$ is height, $\mathrm{v}(t) = \mathrm{h}(t)$ is velocity, $\mathrm{a}(t) = \mathrm{v}(t) = \ddot{\mathrm{h}}(t)$ is acceleration

$$v0 = 120 \frac{\text{feet}}{s}, \quad g = 9,8 \frac{\mathrm{m}}{\mathrm{s}^{2}} = 9,8 \cdot 3,28 \left(\frac{\text{feet}}{\mathrm{s}^{2}}\right) = 32,14 \frac{\text{feet}}{\mathrm{s}^{2}}$$

As we know $\mathrm{a}(t) = \mathrm{const} = -\mathrm{g}$, so $\mathrm{v}(t) = \int \mathrm{a}(t) \, \mathrm{d}t = \mathrm{v}_0 - \mathrm{g}t$ ($\mathrm{v}_0$ is the initial velocity)

Also $\mathrm{h}(t) = \int \mathrm{v}(t) \, \mathrm{d}t = \int (\mathrm{v}_0 - \mathrm{g}t) \, \mathrm{d}t = \mathrm{h}_0 + \mathrm{v}_0 t - \frac{\mathrm{g}t^2}{2}$ ($\mathrm{h}_0$ is the initial height)

$\mathrm{h}(t_{\max}) = \mathrm{h}_{\max}$ when $\dot{\mathrm{h}}(t_{\max}) = \mathrm{v}(t_{\max}) = 0$, so $0 = \mathrm{v}_0 - \mathrm{g}t_{\max}, t_{\max} = \frac{\mathrm{v}_0}{\mathrm{g}}$,

$\mathrm{h}_{\max} = \mathrm{h}(t_{\max}) = \mathrm{h}_0 + \mathrm{v}_0 t - \left(\frac{\mathrm{g}}{2}\right) \left(\frac{\mathrm{v}_0}{\mathrm{g}}\right)^2 = \mathrm{h}_0 + \frac{\mathrm{v}_0^2}{2\mathrm{g}}$ (you can get this much easier, if you know The Law of Conservation of Energy: $\mathrm{mv}^2 + \mathrm{mg}\Delta \mathrm{h} = \mathrm{const}$, so in this case $\mathrm{mv}^2 = \mathrm{mg}\Delta \mathrm{h}, \Delta \mathrm{h} = \frac{\mathrm{v}^2}{2\mathrm{g}}$)

$$\mathrm{h}_{\max} = 2 + \frac{120^2}{2 \cdot 32,14} = 226,02 \text{ (feet)}$$

Answer: 226 feet

5. A tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 110 feet per second. After how many seconds does the ball attain its maximum height? Round to the nearest hundredth.

Solution:

$\mathrm{h}(t)$ is height, $\mathrm{v}(t) = \dot{\mathrm{h}}(t)$ is velocity, $\mathrm{a}(t) = \dot{\mathrm{v}}(t) = \ddot{\mathrm{h}}(t)$ is acceleration

$$v 0 = 1 1 0 \frac {\text {f e e t}}{s}, \quad g = 9, 8 \frac {m}{s ^ {2}} = 9, 8 * 3, 2 8 \left(\frac {\text {f e e t}}{s ^ {2}}\right) = 3 2, 1 4 \frac {\text {f e e t}}{s ^ {2}}$$

As we know $a(t) = \text{const} = -g$ , so $v(t) = \int a(t) dt = v_0 - gt$ ( $v_0$ is the initial condition)

Also $h(t) = \int v(t) \, dt = \int (v_0 - gt) \, dt = h_0 + v_0 t - \frac{gt^2}{2}$

$\mathrm{h}(t_{\max}) = \mathrm{h}_{\max}$ when $\dot{\mathrm{h}} (t_{\max}) = \mathrm{v}(t_{\max}) = 0, \mathrm{so} 0 = \mathrm{v}_0 - \mathrm{gt}_{\max}, t_{\max} = \frac{\mathrm{v}_0}{\mathrm{g}},$

$$t _ {\max } = \frac {1 1 0}{3 2 , 1 4} = 3, 4 2 2 5 (s)$$

Answer: 3,42 s.

6. The finishing time for a runner completing the 200-meter dash is affected by the tail-wind speed, s. The change, t, in a runner's performance is modeled by the function shown below:

$$t = 0. 0 1 1 9 s ^ {2} - 0. 3 0 8 s - 0. 0 0 0 3$$

Predict the change in a runner's finishing time with a wind speed of 5 meters/second. Note: A negative answer means the runner finishes with a lower time. Round to the nearest hundredths.

Solution:

$$t(s) = 0.0119s^2 - 0.308s - 0.0003$$ $$t(5) = 0.0119 * (5^2) - 0.308 * (5) - 0.0003 = -1,2428 \, (s)$$

Answer: -1,24 s, the runner finishes with a lower time.

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