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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 speed is 49 m/s and the height of the bridge is 122.5 meters.

Step by step solution

01

Understand the problem

We are given a scenario where a marble is dropped from a bridge and it takes 5 seconds to hit the water. We need to find the speed at which it strikes the water and the height of the bridge.
02

Solve for final velocity (speed)

Since the marble is dropped, its initial velocity is 0. We use the equation for final velocity in free fall: \( v = v_0 + gt \), where \( v_0 = 0 \), \( g = 9.8 \text{ m/s}^2 \), and \( t = 5 \text{ s} \).So, \( v = 0 + 9.8 \cdot 5 = 49 \text{ m/s} \).This is the speed with which the marble strikes the water.
03

Solve for the height of the bridge

To find the height, we use the equation: \( h = v_0 t + \frac{1}{2}gt^2 \). Again, \( v_0 = 0 \), \( g = 9.8 \text{ m/s}^2 \), and \( t = 5 \text{ s} \).Thus, \( h = 0 \cdot 5 + \frac{1}{2} \cdot 9.8 \cdot 5^2 = \frac{1}{2} \cdot 9.8 \cdot 25 = 122.5 \text{ meters} \).This is the height of the bridge.

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Key Concepts

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

Free Fall
Free fall describes the motion of a body where gravity is the only force acting upon it. When objects are dropped, they accelerate towards the Earth without any initial push or force other than gravity. Gravity on Earth provides an acceleration of approximately \(9.8 \, \text{m/s}^2\).

This means that for every second an object is in free fall, its speed increases by 9.8 meters per second. During free fall, objects do not encounter air resistance or any other forces. This makes calculations simpler since only the constant acceleration due to gravity is involved.
  • Initial velocity for dropped objects is zero.
  • Gravity provides constant acceleration.
  • Air resistance is typically ignored in basic free fall scenarios.
Final Velocity
The final velocity of an object in free fall is the speed it reaches right before it hits the ground. Starting from rest, the final velocity can be calculated using the equation:
\[ v = v_0 + gt \]
where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time the object spends falling.

In the exercise, the marble's initial velocity \(v_0\) is zero because it is dropped. With \(g = 9.8 \, \text{m/s}^2\) and \(t = 5 \, \text{s}\), the final velocity is:
\[ v = 0 + 9.8 \times 5 = 49 \, \text{m/s} \]
Thus, the marble strikes the water at 49 meters per second.

Final velocity provides insight into how fast an object is moving just before impact, a crucial component in understanding motion.
Height Calculation
Calculating the height from which an object falls involves understanding how distance is covered during free fall. The height or distance, \(h\), can be determined with the equation:
\[ h = v_0t + \frac{1}{2}gt^2 \]
In scenarios like our exercise, since the object is dropped, \(v_0\) is zero. This simplifies the equation to only account for the distance covered due to gravity's acceleration:
\[ h = \frac{1}{2} \cdot 9.8 \cdot 5^2 \]
Calculating gives us:
\[ h = \frac{1}{2} \cdot 9.8 \cdot 25 = 122.5 \text{ meters} \]
Thus, the bridge is 122.5 meters high.

Knowing the height helps visualize and measure the extent of the motion, a key element in physics studies.
Equations of Motion
Equations of motion are essential tools in kinematics. They allow us to predict and describe the motion of objects. For free fall scenarios, there are three primary equations to use:
  • \( v = v_0 + gt \) for calculating final velocity.
  • \( h = v_0t + \frac{1}{2}gt^2 \) for determining height.
  • \( v^2 = v_0^2 + 2gh \) which can be used to find any of the variables if others are known.

These equations are based on the principles of constant acceleration. When applying them, it’s crucial that assumptions like negligible air resistance hold true.

By understanding these equations, we can solve a wide range of problems involving motion, whether it is a marble from our example or any other object moving under gravity's influence.

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Most popular questions from this chapter

A baseball is thrown with an initial velocity of \(100 \mathrm{~m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal, as seen in Fig. \(2-6 .\) How far from the throwing point will the baseball attain its original level? Divide the problem into horizontal and vertical parts, for which $$v_{i x}=v_{i} \cos 30.0^{\circ}=86.6 \mathrm{~m} / \mathrm{s} \quad \text { and } \quad v_{i y}=v_{i} \sin 30.0^{\circ}=50.0 \mathrm{~m} / \mathrm{s}$$ where \(u p\) is taken as positive. In the vertical piece of the problem, \(y=0\), since the ball returns to its original height. Then $$y=v_{i y} t+\frac{1}{2} a_{y} t^{2} \quad \text { or } \quad 0=(50.0 \mathrm{~m} / \mathrm{s}) t+\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}$$ therefore, \(\quad-50.0 \mathrm{~m} / \mathrm{s}=\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) t\) and \(t=10.2 \mathrm{~s}\). In the horizontal part of the problem, \(v_{i x}=v_{f x}=v_{x}=86.6 \mathrm{~m} / \mathrm{s}\). Therefore, $$ x=v_{x} t=(86.6 \mathrm{~m} / \mathrm{s})(10.2 \mathrm{~s})=883 \mathrm{~m}. $$

A stone is thrown straight downward with initial speed \(8.0 \mathrm{~m} / \mathrm{s}\) from a height of \(25 \mathrm{~m}\). Find \((a)\) the time it takes to reach the ground and \((b)\) the speed with which it strikes.

A truck's speed increases uniformly from \(15 \mathrm{~km} / \mathrm{h}\) to \(60 \mathrm{~km} / \mathrm{h}\) in \(20 \mathrm{~s}\). Determine \((a)\) the average speed, \((b)\) the acceleration, and \((c)\) the distance traveled, all in units of meters and seconds. For the 20-s trip under discussion, taking \(+x\) to be in the direction of motion, $$\begin{array}{l} v_{i x}=\left(15 \frac{\mathrm{km}}{\mathrm{K}}\right)\left(1000 \frac{\mathrm{m}}{\mathrm{k} \mathrm{m}}\right)\left(\frac{1}{3600} \frac{\mathrm{K}}{\mathrm{s}}\right)=4.17 \mathrm{~m} / \mathrm{s} \\ v_{f x}=60 \mathrm{~km} / \mathrm{h}=16.7 \mathrm{~m} / \mathrm{s} \end{array}$$ (a) \(v_{a v}=\frac{1}{2}\left(v_{i x}+v_{f x}\right)=\frac{1}{2}(4.17+16.7) \mathrm{m} / \mathrm{s}=10 \mathrm{~m} / \mathrm{s}\) (b) \(\quad a=\frac{v_{f x}-v_{i x}}{t}=\frac{(16.7-4.17) \mathrm{m} / \mathrm{s}}{20 \mathrm{~s}}=0.63 \mathrm{~m} / \mathrm{s}^{2}\) (c) \(\quad x=v_{a v} t=(10.4 \mathrm{~m} / \mathrm{s})(20 \mathrm{~s})=208 \mathrm{~m}=0.21 \mathrm{~km}\)

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 \(u p\) as positive. How would the calculation have been changed?)

A stunt flier is moving at \(15 \mathrm{~m} / \mathrm{s}\) parallel to the flat ground \(100 \mathrm{~m}\) below, as illustrated in Fig. \(2-5 .\) How large must the distance \(x\) from plane to target be if a sack of flour released from the plane is to strike the target? Following the same procedure as in Problem \(2.17\), we use \(y=v_{i y} t+\frac{1}{2} a_{y} t^{2}\) to get $$-100 \mathrm{~m}=0+\frac{1}{2}\left(-9.81 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \quad \text { or } \quad t=4.52 \mathrm{~s}$$ Now \(x=v_{x} t=(15 \mathrm{~m} / \mathrm{s})(4.52 \mathrm{~s})=67.8 \mathrm{~m}\) or \(68 \mathrm{~m}\).

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