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A ball is thrown directly downward, with an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , from a height of 30.0 \(\mathrm{m}\) . After what time interval does the ball strike the ground?

Short Answer

Expert verified
The ball strikes the ground after approximately 2.02 seconds.

Step by step solution

01

Identify Given Information

The ball is thrown downward with an initial speed of 8.00 m/s from a height of 30.0 m. The acceleration due to gravity (g) is 9.8 m/s^2, which will be positive since we're taking downward as the positive direction.
02

Set Up the Equation

Use the kinematic equation for uniformly accelerated motion: \( s = ut + \frac{1}{2}at^2 \) where: s = displacement (30.0 m), u = initial velocity (8.00 m/s), a = acceleration due to gravity (9.8 m/s^2), t = time (unknown).
03

Solve for Time (t)

Plug in the known values and solve the quadratic equation: \( 30.0 m = (8.00 m/s) t + \frac{1}{2}(9.8 m/s^2) t^2 \) Simplify the equation to standard quadratic form and solve for t using the quadratic formula where necessary.
04

Find the Solution Using Quadratic Formula

The quadratic formula, \( t = \frac{-b \text{ }}{2a} \text{ }( b^2 - 4ac )^{\frac{1}{2}}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ ) \ 2a }\text{ }\text{ },\) can be used after substituting a = 4.9 m/s^2 (half the acceleration), b = 8.00 m/s (initial velocity), and c = -30.0 m (the negative displacement).
05

Perform the Calculation

Substitute the values into the quadratic formula and solve for t. Discard the negative value of t as time cannot be negative in this context.

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Key Concepts

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

Kinematic Equations
In motion with constant acceleration, kinematic equations enable us to predict an object's future position, velocity, and the time it takes to alter its state of motion.
Kinematic equations are especially helpful in problems where an object is moving under the constant influence of gravity and we need to find variables like displacement, initial velocity, acceleration, or time. They provide a clear way to relate these variables and solve for unknowns.
One of the four key kinematic equations is:
\[ s = ut + \frac{1}{2}at^2 \]
where:
  • \( s \) is the displacement, the change in position of the object,
  • \( u \) is the initial velocity, the starting speed in a specified direction,
  • \( a \) is the constant acceleration, and
  • \( t \) is the time over which the motion occurs.
In the given exercise, we apply this equation to determine the time it takes for a ball thrown downward to reach the ground. Through a proper substitution of the known values, we are led to the need for solving a quadratic equation to find the time, \( t \).
Quadratic Formula
When faced with a kinematic equation that has been rearranged into a quadratic form, the quadratic formula is an invaluable mathematical tool for finding the unknown variable.
The general form of a quadratic equation is:
\[ ax^2 + bx + c = 0 \]
The quadratic formula that solves for \( x \) is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.
The '±' symbol indicates that there will generally be two solutions. In kinematic problems, we often need to discard one solution (for example, a negative time), because it may not be physically meaningful in the context of the problem.
In our sample problem, the kinematic equation was manipulated into a quadratic equation of time \( t \). By substituting the corresponding coefficients into the quadratic formula, we obtain the time at which the ball will hit the ground. Remember to consider the physical context and choose the solution that makes sense—in this case, the positive value for time.
Projectile Motion
Projectile motion is a form of motion where an object moves in a curved trajectory under the influence of gravity, and it's an important application of kinematic equations. To solve problems of projectile motion, we generally break the motion into horizontal and vertical components.
In the textbook problem, a ball is thrown downward, which is a one-dimensional case of projectile motion where the horizontal displacement is not considered, and gravity acts as the sole acceleration.
In a more complex scenario where an object is projected at an angle, the kinematic equations must be used separately for the horizontal motion (where there's no acceleration) and the vertical motion (where acceleration due to gravity is present).
Understanding projectile motion is vital to predicting the path of a thrown object, how long it will be in motion, and how far it will travel. This concept has both theoretical and practical applications across various fields such as sports, engineering, and physics.

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Most popular questions from this chapter

A speedboat moving at 30.0 \(\mathrm{m} / \mathrm{s}\) approaches a no-wake buoy marker 100 \(\mathrm{m}\) ahead. The pilot slows the boat with a constant acceleration of \(-3.50 \mathrm{m} / \mathrm{s}^{2}\) by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy?

A motorist drives along a straight road at a constant speed of 15.0 \(\mathrm{m} / \mathrm{s}\) . Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 2.00 \(\mathrm{m} / \mathrm{s}^{2}\) to over- take her. Assuming the officer maintains this acceleration, (a) determine the time it takes the police officer to reach the motorist. Find \((\mathrm{b})\) the speed and \((\mathrm{c})\) the total displacement of the officer as he overtakes the motorist.

A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.0 \(\mathrm{m} / \mathrm{s}\) at ground level. Its engines then fire and it accelerates upward at 4.00 \(\mathrm{m} / \mathrm{s}^{2}\) until it reaches an altitude of 1000 \(\mathrm{m}\) . At that point its engines fail and the rocket goes into free fall, with an acceleration of \(-9.80 \mathrm{m} / \mathrm{s}^{2}\) . (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (You will need to consider the motion while the engine is operating separate from the free-fall motion.)

It is possible to shoot an arrow at a speed as high as 100 \(\mathrm{m} / \mathrm{s}\) . (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?

Jules Verne in 1865 suggested sending people to the Moon by firing a space capsule from a 220 -m-long cannon with a launch speed of 10.97 \(\mathrm{km} / \mathrm{s}\) . What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration 9.80 \(\mathrm{m} / \mathrm{s}^{2}\) .

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