Questions: 3. If your friend drops a chocolate bar to you from a height of 5.0 m above your hands, (a) How long would it take you to catch it? Use Eq. 2. height = 5.0 m initial (0, ? ) h = V1 t + 1 / 2 g = 9.8 m / s^2 t = sqrt(2 h / 9) t = sqrt(2 * 5.0 m) / 9.8 m / s^2 ≈ 1.0102 s t = sqrt(1.8 / 9 .8) t = sqrt(1.0204) (b) With what speed will the chocolate bar strike your hands? Use Eq. 3. g = 9.8 m / s^2 ≈ 1.0102 s h = 5.0 m v = g t → 9.9 m / s^2 × 1.01025 ≈ 9.8996 mb (c) In the previous part, what would the speed equal if you had incorrectly used Eq. 1 instead? (d) Why is using Eq. 1 incorrect in this problem?
Transcript text: 3. If your friend drops a chocolate bar to you from a height of 5.0 m above your hands, (a) How long would it take you to catch it? Use Eq. 2.
\[
\begin{array}{l}
\text { height }=5.0 \mathrm{~m} \\
\text { initial }(0, \text { ? }) \\
h=V_{1} t+1 / 2 \\
g=9.8 \mathrm{~m} / \mathrm{s}^{2} \\
t=\sqrt{\frac{2 h}{9}} \\
t=\frac{\sqrt{2 \times 5.0 \mathrm{~m}}}{9.8 \mathrm{~m} / \mathrm{s}^{2}} \\
\approx 1.0102 \mathrm{~s} \\
t=\sqrt{\frac{1.8}{9} .8} \quad t=\sqrt{1.0204}
\end{array}
\]
(b) With what speed will the chocolate bar strike your hands? Use Eq. 3.
\[
\begin{array}{l}
g=9.8 \mathrm{~m} / \mathrm{s}^{2} \approx 1.0102 \mathrm{~s} \\
h=5.0 \mathrm{~m} \\
v=g t \rightarrow 9.9 \mathrm{~m} / \mathrm{s}^{2} \times 1.01025 \\
\approx 9.8996 \mathrm{mb}
\end{array}
\]
(c) In the previous part, what would the speed equal if you had incorrectly used Eq. 1 instead?
(d) Why is using Eq. 1 incorrect in this problem?
Solution
Was this solution helpful?
Answered
