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A baseball is thrown straight upward with a speed of \(30 \mathrm{~m} / \mathrm{s}\). (a) How long will it rise? ( \(b\) ) How high will it rise? (c) How long after it leaves the hand will it return to the starting point? ( \(d\) ) When will its speed be \(16 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
(a) 3.06 s; (b) 45.92 m; (c) 6.12 s; (d) 4.69 s.

Step by step solution

01

Understanding the given data

We know the initial speed of the baseball is given as \( v_i = 30 \, \text{m/s} \). The acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \) directed downwards.
02

Calculate the time to reach the highest point (Part a)

At the highest point, the vertical velocity of the baseball is \(0\). We use the equation: \( v_f = v_i - g \cdot t \). Setting \( v_f = 0 \), we have: \[ 0 = 30 - 9.8t \] Solving for \( t \), \[ t = \frac{30}{9.8} \approx 3.06 \, \text{s} \]
03

Calculate the maximum height reached (Part b)

Use the kinematic equation: \( v_f^2 = v_i^2 - 2g \cdot h \). Substituting \( v_f = 0 \), we have: \[ 0 = 30^2 - 2 \cdot 9.8 \cdot h \] \[ h = \frac{30^2}{2 \cdot 9.8} \approx 45.92 \, \text{m} \]
04

Determine the total time of flight (Part c)

The total time of flight is twice the time to reach the maximum height since the upward and downward journeys are symmetric (ignoring air resistance). From Step 2, the time to rise is about \( 3.06 \, \text{s} \), so total time is: \[ T = 2 \times 3.06 = 6.12 \, \text{s} \]
05

Find the time to reach 16 m/s speed on the way down (Part d)

On the way down, when the speed is \( 16 \, \text{m/s} \), the equation is \( v = g \cdot t \). Solving for \( t \): \[ 16 = 9.8t \rightarrow t = \frac{16}{9.8} \approx 1.63 \, \text{s} \]Adding this to the time to reach the top gives \( 3.06 + 1.63 \approx 4.69 \, \text{s} \) after leaving the hand.

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Key Concepts

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

Kinematic Equations
Kinematic equations are fundamental in physics, especially for analyzing motion. They describe the motion of objects under the influence of gravity and initial velocity. For projectile motion, where an object is propelled in space subjected to gravitational force, kinematic equations come in handy. They help us determine quantities like time, displacement, final velocity, and acceleration.

For instance, to calculate how long a baseball will rise, one could start with the equation for final velocity:
  • \[ v_f = v_i - g \cdot t \]
Here, \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( g \) is the acceleration due to gravity, and \( t \) is time. The beauty of these equations is that they relate various properties of the motion, enabling us to solve for unknown quantities with the available data. Other commonly used equations include those for displacement and energy conservation, like:
  • \[ v_f^2 = v_i^2 - 2g \cdot h \]
Acceleration due to Gravity
Acceleration due to gravity is a vital concept in projectile motion. It is the constant force that acts on an object in freefall towards the center of the Earth. Its standard value on Earth's surface is approximately
  • \( g = 9.8 \, \text{m/s}^2 \)
This acceleration is responsible for reducing a projectile's vertical velocity as it ascends, bringing it momentarily to zero at its highest point, and then increasing its velocity on descent.

Understanding this constant allows us to predict how long and how high an object will rise. For example, when a baseball is thrown upwards, its speed decreases by 9.8 m/s every second it travels upwards due to gravity. Thus, it will rise until gravity uses up all its initial kinetic energy.
Time of Flight
Time of flight refers to the total time an object remains in the air from the moment it is launched until it returns to its starting point. In projectile motion without air resistance, this time is symmetric—the time taken to rise to the peak equals the time to descend. This property can simplify calculations significantly.

For instance, if a baseball takes about 3.06 seconds to reach its maximum height, the total time of flight will be:
  • \[ T = 2 \times 3.06 \approx 6.12 \, \text{s} \]
The symmetry comes from the constant deceleration during the ascent and constant acceleration during the descent caused by gravity. Thus, knowing the time to the vertex helps in calculating the full trajectory without complexity.
Vertical Velocity
Vertical velocity pertains to the speed of an object moving along the vertical axis. When analyzing projectile motion, tracking vertical velocity reveals the effects of gravitational acceleration over time.
  • Initially, a projectile's vertical velocity is equal to its launch speed in the upward direction.
  • As the projectile rises, gravity decreases this velocity until it becomes zero at the peak.
  • Upon descent, the vertical velocity increases again but in the downward direction.
To solve when a baseball's speed reaches 16 m/s on its way down, utilizing the relation \( v = g \cdot t \) allows calculation of the precise time taken for this velocity change. By solving such problems, students gain a deeper understanding of motion principles, reinforcing concepts such as conservation of energy and constant acceleration.

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Most popular questions from this chapter

A body with initial velocity \(8.0 \mathrm{~m} / \mathrm{s}\) moves along a straight line with constant positive acceleration and travels \(640 \mathrm{~m}\) in \(40 \mathrm{~s}\). For the 40 s interval, find \((a)\) the average velocity, \((b)\) the final velocity, and \((c)\) the acceleration.

A body falls freely from rest. Find \((a)\) its acceleration, \((b)\) the distance it falls in \(3.0 \mathrm{~s},(c)\) its speed after falling \(70 \mathrm{~m},(d)\) the time required to reach a speed of \(25 \mathrm{~m} / \mathrm{s}\), and \((e)\) the time taken to fall \(300 \mathrm{~m}\).

A nut comes loose from a bolt on the bottom of an elevator as the elevator is moving up the shaft at \(3.00 \mathrm{~m} / \mathrm{s}\). The nut strikes the bottom of the shaft in \(2.00\) s. \((a)\) How far from the bottom of the shaft was the elevator when the nut fell off? (b) How far above the bottom was the nut \(0.25\) s after it fell off?

(a) Find the range \(x\) of a gun that fires a shell with muzzle velocity \(u\) at an angle of elevation \(\theta\). ( \(b\) ) Find the angle of elevation \(\theta\) of a gun that fires a shell with a muzzle velocity of \(120 \mathrm{~m} / \mathrm{s}\) and hits a target on the same level but \(1300 \mathrm{~m}\) distant. (See \(\underline{\text { Fig. } 2-8 .}\) ) (a) Let \(t\) be the time it takes the shell to hit the target. Then, \(x=v_{i x}\) tor \(t=x / v_{i x}\). Consider the vertical motion alone, and take \(u p\) as positive. When the shell strikes the target, $$ \text { Vertical displacement }=0=v_{i y} t+\frac{1}{2}(-g) t^{2} $$ Solving this equation gives \(t=2 v_{i y} / g .\) But \(t=x / v_{i x}\), so $$ \frac{x}{v_{i x}}=\frac{2 v_{i y}}{g} \quad \text { or } \quad x=\frac{2 v_{i x} v_{i y}}{g}=\frac{2\left(v_{i} \cos \theta\right)\left(v_{i} \sin \theta\right)}{g} $$ wherein \(g\) is positive. The formula \(2 \sin \theta \cos \theta=\sin 2 \theta\) can be used to simplify this. After substitution, $$ x=\frac{v_{i}^{2} \sin 2 \theta}{g} \text { . } $$ The maximum range corresponds to \(\theta=45^{\circ}\), since \(\sin 2 \theta\) has a maximum value of 1 when \(2 \theta=90^{\circ}\) or \(\theta=45^{\circ}\). (b) From the range equation found in \((a)\), $$ \sin 2 \theta=\frac{g x}{v_{i}^{2}}=\frac{\left(9.18 \mathrm{~m} / \mathrm{s}^{2}\right)(1300 \mathrm{~m})}{(120 \mathrm{~m} / \mathrm{s})^{2}}=0.886 $$ Therefore, \(2 \theta=\arcsin 0.886=62^{\circ}\) and so \(\theta=31^{\circ}\).

People working for National Geographic dropped a peregrine falcon from a plane at an altitude of \(4572 \mathrm{~m}(15000 \mathrm{ft})\). The bird dove down reaching a speed of about \(81.8 \mathrm{~m} / \mathrm{s}(183 \mathrm{mph})\). Determine its acceleration assuming it to be constant. [Hint: The bird was dropped, not thrown down.]

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