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Chapter 2: Problem 37

A marble dropped from a bridge strikes the water in \(5.0 \mathrm{~s}\). Calculate ( \(a\) ) the speed with which it strikes and \((b)\) the height of the bridge.

Short Answer

Expert verified
The marble strikes the water at 49 m/s and the height of the bridge is 122.5 m.

Step by step solution

01

Identify Known Values

We know the time the marble takes to fall is \( t = 5.0 \text{ s} \). We'll assume the initial velocity \( u = 0 \) since the marble is dropped (not thrown). The acceleration due to gravity \( g \) is approximately \( 9.8 \text{ m/s}^2 \).
02

Calculate Final Speed

To find the speed with which the marble strikes the water, use the formula for final velocity in free fall: \[ v = u + gt \]Substitute the known values: \[ v = 0 + (9.8 \text{ m/s}^2)(5.0 \text{ s}) = 49.0 \text{ m/s} \] Therefore, the speed when it strikes the water is 49 m/s.
03

Calculate Height of the Bridge

To find the height from which the marble was dropped, use the formula for distance covered:\[ h = ut + \frac{1}{2}gt^2 \]Substitute the known values:\[ h = 0 \times 5.0 + \frac{1}{2}(9.8 \text{ m/s}^2)(5.0 \text{ s})^2 \]\[ h = \frac{1}{2} \times 9.8 \times 25 = 122.5 \text{ m} \] So, the height of the bridge is 122.5 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Acceleration Due to Gravity
When objects are in free fall, they move under the influence of gravity alone. On Earth, the force of gravity gives any falling object an acceleration of approximately \( 9.8 \text{ m/s}^2\). This is known as the acceleration due to gravity.
Imagine you drop a marble from a height鈥攂ecause of gravity, it will accelerate downwards, increasing its velocity every second until it hits the ground. Notably:
  • Constant Rate: This acceleration remains constant, meaning each second the velocity of the object increases by \( 9.8 \text{ m/s}\).
  • Independent of Mass: Surprisingly, this acceleration is the same for a heavy marble or a light feather (in the absence of air resistance), indicating that mass doesn't affect the acceleration.
Knowing this constant acceleration helps us calculate other parameters, such as the velocity or distance a falling object travels over time.
Calculating Velocity in Free Fall
Velocity indicates how fast something moves in a particular direction. In the context of free fall, finding the velocity at which an object hits the ground involves a simple calculation:
Since the marble was dropped (initial velocity \( u = 0 \)), its velocity increases steadily due to gravitational acceleration. To calculate the final velocity \( v \), use the formula:\[ v = u + gt \]
  • Initial Velocity \( u \): At the start, the marble has no initial velocity (\( 0 \text{ m/s}\)) as it is simply released.
  • Time \( t \): This is how long the marble has been falling. In your problem, \( t = 5.0 \text{ s}\).
  • Gravitational Acceleration \( g \): Using the value \( 9.8 \text{ m/s}^2 \).
  • Calculation Outcome: Substitute these into the formula: \[ v = 0 + 9.8 \times 5 = 49 \text{ m/s}\]. Thus, the marble is moving at 49 m/s when it reaches the water.
Understanding this can help predict how quickly other objects reach the ground when dropped from a similar height.
Determining Distance Fallen
Calculating how far a marble falls involves using a specific formula that accounts for both the initial velocity and the time it falls. Since in this scenario the marble was dropped (initial velocity is zero), the distance formula simplifies nicely.
The height from which it was dropped, or the distance \( h \) it falls, can be calculated using: \[ h = ut + \frac{1}{2}gt^2 \]
  • Initial Velocity \( u \): Again, this is \( 0 \text{ m/s} \) since the drop starts from rest.
  • Time \( t \): The time during which gravity acts on the marble, \( 5.0 \text{ s}\) in this case.
  • Gravitational Acceleration \( g \): Consistently \( 9.8 \text{ m/s}^2\).
  • Formula Application: Plugging these into the equation: \[ h = 0 \times 5.0 + \frac{1}{2} \times 9.8 \times (5.0^2) = 122.5 \text{ m}\]. Therefore, the marble falls from a height of 122.5 meters.
Knowing how to calculate this distance is useful for understanding how gravity affects falling objects, especially when they start from rest.

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Most popular questions from this chapter

A stone is thrown straight upward with a speed of \(20 \mathrm{~m} / \mathrm{s}\). It is caught on its way down at a point \(5.0 \mathrm{~m}\) above where it was thrown. (a) How fast was it going when it was caught? ( \(b\) ) How long did the trip take? The situation is shown in Fig. 2-3. Take up as positive. Then, for the trip that lasts from the instant after throwing to the instant before catching, \(v_{i y}=20 \mathrm{~m} / \mathrm{s}, y=+5.0 \mathrm{~m}\) (since it is an upward displacement), \(a=-9.81 \mathrm{~m} / \mathrm{s}^{2}\) (a) Use \(v_{f y}^{2}=v_{i y}^{2}+2 a y\) to compute $$ \begin{array}{l} v_{f j}^{2}=(20 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(5.0 \mathrm{~m})=302 \mathrm{~m}^{2} / \mathrm{s}^{2} \\\ v_{f j}=\pm \sqrt{302 \mathrm{~m}^{2} / \mathrm{s}^{2}}=-17 \mathrm{~m} / \mathrm{s} \end{array} $$ Take the negative sign because the stone is moving downward, in the negative direction, at the final instant. (b) To find the time, use \(a=\left(v_{f y}-v_{i y}\right) / t\) and so $$ t=\frac{(-17.4-20) \mathrm{m} / \mathrm{s}}{-9.81 \mathrm{~m} / \mathrm{s}^{2}}=3.8 \mathrm{~s} $$ Notice that we retain the minus sign on \(v_{f y}\). You can check your work by dividing the problem into two parts, the trip up to peak altitude and the trip down from peak. The peak altitude is given by Eq. (2.9); that is, \(y_{p}=-v_{i}^{2} / 2 g=-(20\) \(\mathrm{m} / \mathrm{s})^{2} / 2\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=20.38736 \mathrm{~m}\). [Hint: Don't round off to two figures mid-calculation.] Now drop the stone from \(y_{p}\) so it falls a distance \((20.38736 \mathrm{~m})-(5.0 \mathrm{~m})=15.38736 \mathrm{~m}\), at which point it will be moving- - from Eq. (2.6) with down as plus-at \(v_{f}^{2}=2 g s=\) \(2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(15.38736 \mathrm{~m})=301.900 \mathrm{~m}^{2} / \mathrm{s}^{2}\), and so \(v_{f}=17.4 \mathrm{~m} / \mathrm{s}=\) \(17 \mathrm{~m} / \mathrm{s}\). Similarly, you can calculate the total time of flight, which equals the time to reach peak altitude plus the time to fall \(15.387\) \(\mathrm{m} .\)

A baseball is thrown straight upward on the Moon with an initial speed of \(35 \mathrm{~m} / \mathrm{s}\). Compute \((a)\) the maximum height reached by the ball, (b) the time taken to reach that height, \((c)\) its velocity \(30 \mathrm{~s}\) after it is thrown, and \((d)\) when the ball's height is \(100 \mathrm{~m}\). Take up as positive. At the highest point, the ball's velocity is zero. (a) From \(v_{f y}^{2}=v_{i y}^{2}+2 a y\), since \(g=1.60 \mathrm{~m} / \mathrm{s}^{2}\) on the Moon, $$ 0=(35 \mathrm{~m} / \mathrm{s})^{2}+2\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) y \quad \text { or } \quad y=0.38 \mathrm{~km} $$ (b) From \(v_{f y}=v_{i y}+a t\) $$ 0=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t \quad \text { or } \quad t=22 \mathrm{~s} $$ (c) From \(v_{f y}=v_{i y}+\) at $$ v_{f j}=35 \mathrm{~m} / \mathrm{s}+\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right)(30 \mathrm{~s}) \quad \text { or } \quad v_{f y}=-13 \mathrm{~m} / \mathrm{s} $$ Because \(v_{f}\) is negative and we are taking up as positive, the velocity is directed downward. The ball is on its way down at \(t\) \(=30 \mathrm{~s} .\) (d) From \(y=v_{i y} t+\frac{1}{2} a t^{2}\) we have $$ 100 \mathrm{~m}=(34 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-1.60 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \quad \text { or } \quad 0.80 t^{2}-35 t+100=0 $$ By use of the quadratic formula, Or $$ \begin{aligned} x &=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ t &=\frac{35 \pm \sqrt{35^{2}-4(0.80) 100}}{2(0.80)}=\frac{35 \pm 30.08}{1.60} \end{aligned} $$ we find \(t=3.1 \mathrm{~s}\) and \(41 \mathrm{~s}\). At \(t=3.1 \mathrm{~s}\) the ball is at \(100 \mathrm{~m}\) and ascending; at \(t=41 \mathrm{~s}\) it is at the same height but descending.

A ball is dropped from rest at a height of \(50 \mathrm{~m}\) above the ground. (a) What is its speed just before it hits the ground? (b) How long does it take to reach the ground? If we can ignore air friction, the ball is uniformly accelerated until it reaches the ground. Its acceleration is downward and is \(9.81\) \(\mathrm{m} / \mathrm{s}^{2}\). Taking down as positive, we have for the trip: $$ y=50.0 \mathrm{~m} \quad a=9.81 \mathrm{~m} / \mathrm{s}^{2} \quad v_{i}=0 $$ (a) \(v_{f y}^{2}=v_{i y}^{2}+2 a y=0+2\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(50.0 \mathrm{~m})=981 \mathrm{~m}^{2} / \mathrm{s}^{2}\) and so \(v_{f}=31.3 \mathrm{~m} / \mathrm{s}\). (b) From \(a\left(v_{f y}-v_{i y}\right) / t\), $$ t=\frac{v_{f y}-v_{i y}}{a}=\frac{(31.3-0) \mathrm{m} / \mathrm{s}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=3.19 \mathrm{~s} $$ (We could just as well have taken up as positive. How would the calculation have been changed?)

A marble, rolling with speed \(20 \mathrm{~cm} / \mathrm{s}\), rolls off the edge of a table that is \(80 \mathrm{~cm}\) high. (a) How long does it take to drop to the floor? (b) How far, horizontally, from the table edge does the marble strike the floor?

As drawn in Fig. 2-7, a ball is thrown from the top of one building toward a tall building \(50 \mathrm{~m}\) away. The initial velocity of the ball is \(20 \mathrm{~m} / \mathrm{s}-40^{\circ}\) above horizontal. How far above or below its original level will the ball strike the opposite wall? We have $$ \begin{array}{l} v_{i x}=(20 \mathrm{~m} / \mathrm{s}) \cos 40^{\circ}=15.3 \mathrm{~m} / \mathrm{s} \\ v_{i x}=(20 \mathrm{~m} / \mathrm{s}) \cos 40^{\circ}=12.9 \mathrm{~m} / \mathrm{s} \end{array} $$ Consider first the horizontal motion. For it, $$ v_{i x}=v_{f x}=v_{x} 15.3 \mathrm{~m} / \mathrm{s} $$ Then \(x=v_{x} t\) gives $$ 50 \mathrm{~m}=(15.3 \mathrm{~m} / \mathrm{s}) t \quad \text { or } \quad t=3.27 \mathrm{~s} $$ For the vertical motion, taking down as positive, $$ y=v_{i j} t+\frac{1}{2} a_{y} t^{2}=(-12.9 \mathrm{~m} / \mathrm{s})(3.27 \mathrm{~s})+\frac{1}{2}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.27 \mathrm{~s})^{2}=10.3 \mathrm{~m} $$ and to two significant figures, \(y=10 \mathrm{~m}\). Since \(y\) is positive, and since down is positive, the ball will hit at \(10 \mathrm{~m}\) below the original level.
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