Question

an object is dropped from a height of 100.0 meters. what is the time interval of this fall? (be sure your answer has the appropriate number of significant figures.)

$$\vec { v } _ { f y } = \vec { v } _ { 0 y } + \vec { a } _ { y } \delta t$$

$$\delta \vec { y } = \vec { v } _ { 0 y } \delta t + \frac { 1 } { 2 } \vec { a } _ { y } \delta t ^ { 2 }$$

$$\delta \vec { y } = ( \frac { \vec { v } _ { f y } + \vec { v } _ { 0 y } } { 2 } ) \delta t$$

Explanation:

Step1: Identify known values

The object is dropped, so initial vertical velocity \( \vec{v}_{0y} = 0 \, \text{m/s} \). Displacement \( \Delta \vec{y} = - 100.0 \, \text{m} \) (negative because it's downward), acceleration \( \vec{a}_y = - 9.80 \, \text{m/s}^2 \) (negative for downward direction). We use the equation \( \Delta \vec{y} = \vec{v}_{0y}\Delta t + \frac{1}{2}\vec{a}_y\Delta t^2 \).

Substitute known values: \( -100.0 = 0 \times \Delta t + \frac{1}{2} \times (-9.80) \times \Delta t^2 \)

Step2: Simplify the equation

Simplify to \( -100.0 = -4.90 \times \Delta t^2 \). Multiply both sides by -1: \( 100.0 = 4.90 \times \Delta t^2 \)

Step3: Solve for \( \Delta t^2 \)

\( \Delta t^2 = \frac{100.0}{4.90} \approx 20.408 \)

Step4: Solve for \( \Delta t \)

Take square root: \( \Delta t = \sqrt{20.408} \approx 4.517 \, \text{s} \). Considering significant figures (100.0 has 4, 9.80 has 3, so result should have 3? Wait, 100.0 is 4 sig figs, 9.80 is 3. Wait, the equation: \( \Delta t = \sqrt{\frac{2|\Delta y|}{g}} \), \( |\Delta y| = 100.0 \, \text{m} \), \( g = 9.80 \, \text{m/s}^2 \). So \( \frac{2\times100.0}{9.80} = \frac{200.0}{9.80} \approx 20.408 \), square root is \( \approx 4.517 \), which rounds to 4.52 s? Wait, 100.0 is 4 sig figs, 9.80 is 3. The rule is that when dividing/multiplying, the result has the same number of sig figs as the least precise measurement. But 2 is exact (from the equation). So \( 2\times100.0 = 200.0 \) (4 sig figs), divided by 9.80 (3 sig figs) gives \( \approx 20.408 \) (we keep more digits for intermediate steps), then square root. Wait, maybe the problem expects using \( g = 9.81 \, \text{m/s}^2 \)? Let's check with \( g = 9.81 \): \( \frac{2\times100.0}{9.81} \approx 20.387 \), square root is \( \approx 4.515 \, \text{s} \), so approximately 4.52 s? Wait, but let's do it properly.

Wait, the equation is \( \Delta y = v_{0y}t + \frac{1}{2} a_y t^2 \). Since \( v_{0y} = 0 \), \( t = \sqrt{\frac{2\Delta y}{a_y}} \). But \( \Delta y \) is -100, \( a_y \) is -9.8, so the negatives cancel. So \( t = \sqrt{\frac{2\times100}{9.8}} \approx \sqrt{20.408} \approx 4.517 \, \text{s} \), which is approximately 4.52 s (three significant figures, since 9.8 has two? Wait, 9.80 has three. Wait, the problem says "100.0 meters" which is four significant figures, and if we use \( g = 9.80 \, \text{m/s}^2 \) (three sig figs), then the result should have three sig figs. So 4.52 s? Wait, 4.517 rounds to 4.52. Alternatively, if we use \( g = 9.81 \), it's 4.515, which is 4.52.

Wait, let's recalculate:

\( t = \sqrt{\frac{2h}{g}} \), where \( h = 100.0 \, \text{m} \), \( g = 9.80 \, \text{m/s}^2 \)

\( 2h = 200.0 \, \text{m} \)

\( 200.0 / 9.80 = 20.4081632653 \)

Square root: \( \sqrt{20.4081632653} = 4.51753729 \, \text{s} \), so with three significant figures (from 9.80), it's 4.52 s. But 100.0 has four, 9.80 has three, so the least is three, so 4.52 s.

Answer:

4.52 s