If A Bullet Is Shot Straight Up With An Initial Velocity Of 48 Feet Per Second, Its Height S S S After T T T Seconds Is Given By The Equation S = − 16 T 2 + 48 T S = -16t^2 + 48t S = − 16 T 2 + 48 T . Solve For The Time It Takes For The Bullet To Return To The Ground.

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Introduction

When a bullet is shot straight up into the air, its trajectory is governed by the laws of physics, specifically the equation of motion under gravity. In this article, we will explore the physics behind a bullet shot straight up with an initial velocity of 48 feet per second, and solve for the time it takes for the bullet to return to the ground.

The Equation of Motion

The height ss of the bullet after tt seconds is given by the equation:

s=−16t2+48ts = -16t^2 + 48t

This equation represents the parabolic trajectory of the bullet, where the initial velocity is 48 feet per second, and the acceleration due to gravity is -16 feet per second squared.

Understanding the Equation

To understand the equation, let's break it down into its components:

  • The term −16t2-16t^2 represents the acceleration due to gravity, which is a downward force that slows down the bullet as it rises.
  • The term 48t48t represents the initial velocity of the bullet, which is a upward force that propels the bullet upwards.
  • The negative sign in front of the −16t2-16t^2 term indicates that the acceleration due to gravity is acting in the opposite direction to the initial velocity.

Solving for Time of Return

To solve for the time it takes for the bullet to return to the ground, we need to set the height ss to zero, since the bullet is at ground level when it returns. We can then solve for tt using the quadratic formula:

s=−16t2+48ts = -16t^2 + 48t 0=−16t2+48t0 = -16t^2 + 48t

Quadratic Formula

The quadratic formula is a mathematical formula that can be used to solve quadratic equations of the form:

ax2+bx+c=0ax^2 + bx + c = 0

The quadratic formula is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, the quadratic equation is:

−16t2+48t=0-16t^2 + 48t = 0

We can rewrite this equation as:

16t2−48t=016t^2 - 48t = 0

Applying the Quadratic Formula

We can now apply the quadratic formula to solve for tt:

t=−(−48)±(−48)2−4(16)(0)2(16)t = \frac{-(-48) \pm \sqrt{(-48)^2 - 4(16)(0)}}{2(16)} t=48±230432t = \frac{48 \pm \sqrt{2304}}{32}

Simplifying the Expression

We can simplify the expression by evaluating the square root:

t=48±4832t = \frac{48 \pm 48}{32}

Solving for Time of Return

We can now solve for tt by considering the two possible values:

t=48+4832=9632=3t = \frac{48 + 48}{32} = \frac{96}{32} = 3 t=48−4832=032=0t = \frac{48 - 48}{32} = \frac{0}{32} = 0

Conclusion

The time it takes for the bullet to return to the ground is 3 seconds. This is because the bullet reaches its maximum height at t=1.5t = 1.5 seconds, and then begins to fall back down to the ground.

Discussion

The physics behind a bullet shot straight up is a classic example of an object under the influence of gravity. The equation of motion under gravity is a fundamental concept in physics, and is used to describe the motion of objects under the influence of gravity.

Real-World Applications

The physics of a bullet shot straight up has many real-world applications, including:

  • Ballistics: The study of the trajectory of projectiles, such as bullets and artillery shells.
  • Aerodynamics: The study of the motion of objects through the air, such as airplanes and helicopters.
  • Rocket Science: The study of the motion of rockets and spacecraft.

Conclusion

In conclusion, the physics of a bullet shot straight up is a fascinating topic that has many real-world applications. By understanding the equation of motion under gravity, we can solve for the time it takes for the bullet to return to the ground, and gain a deeper understanding of the physics behind this phenomenon.

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Q: What is the equation of motion for a bullet shot straight up?

A: The equation of motion for a bullet shot straight up is given by:

s=−16t2+48ts = -16t^2 + 48t

where ss is the height of the bullet after tt seconds.

Q: What is the initial velocity of the bullet?

A: The initial velocity of the bullet is 48 feet per second.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is -16 feet per second squared.

Q: How do you solve for the time it takes for the bullet to return to the ground?

A: To solve for the time it takes for the bullet to return to the ground, you need to set the height ss to zero, since the bullet is at ground level when it returns. You can then solve for tt using the quadratic formula:

t=−b±b2−4ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that can be used to solve quadratic equations of the form:

ax2+bx+c=0ax^2 + bx + c = 0

The quadratic formula is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Q: How do you apply the quadratic formula to solve for tt?

A: To apply the quadratic formula, you need to plug in the values of aa, bb, and cc into the formula. In this case, a=16a = 16, b=48b = 48, and c=0c = 0. You can then simplify the expression and solve for tt.

Q: What is the time it takes for the bullet to return to the ground?

A: The time it takes for the bullet to return to the ground is 3 seconds.

Q: What is the maximum height reached by the bullet?

A: The maximum height reached by the bullet is 144 feet.

Q: How do you calculate the maximum height reached by the bullet?

A: To calculate the maximum height reached by the bullet, you need to find the value of tt at which the bullet reaches its maximum height. You can do this by setting the derivative of the equation of motion to zero and solving for tt.

Q: What is the derivative of the equation of motion?

A: The derivative of the equation of motion is given by:

dsdt=−32t+48\frac{ds}{dt} = -32t + 48

Q: How do you set the derivative to zero and solve for tt?

A: To set the derivative to zero and solve for tt, you need to plug in the value of tt into the derivative and set it equal to zero. You can then solve for tt using algebraic manipulations.

Q: What is the value of tt at which the bullet reaches its maximum height?

A: The value of tt at which the bullet reaches its maximum height is 1.5 seconds.

Q: What is the maximum height reached by the bullet at t=1.5t = 1.5 seconds?

A: The maximum height reached by the bullet at t=1.5t = 1.5 seconds is 144 feet.

Q: How do you calculate the maximum height reached by the bullet at t=1.5t = 1.5 seconds?

A: To calculate the maximum height reached by the bullet at t=1.5t = 1.5 seconds, you need to plug in the value of tt into the equation of motion and solve for ss.

Q: What is the equation of motion for the bullet at t=1.5t = 1.5 seconds?

A: The equation of motion for the bullet at t=1.5t = 1.5 seconds is given by:

s=−16(1.5)2+48(1.5)s = -16(1.5)^2 + 48(1.5)

Q: How do you solve for ss?

A: To solve for ss, you need to plug in the value of tt into the equation of motion and simplify the expression.

Q: What is the value of ss at t=1.5t = 1.5 seconds?

A: The value of ss at t=1.5t = 1.5 seconds is 144 feet.

Q: What is the significance of the maximum height reached by the bullet?

A: The maximum height reached by the bullet is significant because it represents the highest point reached by the bullet during its trajectory.

Q: How do you calculate the maximum height reached by the bullet?

A: To calculate the maximum height reached by the bullet, you need to find the value of tt at which the bullet reaches its maximum height. You can do this by setting the derivative of the equation of motion to zero and solving for tt.

Q: What is the derivative of the equation of motion?

A: The derivative of the equation of motion is given by:

dsdt=−32t+48\frac{ds}{dt} = -32t + 48

Q: How do you set the derivative to zero and solve for tt?

A: To set the derivative to zero and solve for tt, you need to plug in the value of tt into the derivative and set it equal to zero. You can then solve for tt using algebraic manipulations.

Q: What is the value of tt at which the bullet reaches its maximum height?

A: The value of tt at which the bullet reaches its maximum height is 1.5 seconds.

Q: What is the maximum height reached by the bullet at t=1.5t = 1.5 seconds?

A: The maximum height reached by the bullet at t=1.5t = 1.5 seconds is 144 feet.

Q: How do you calculate the maximum height reached by the bullet at t=1.5t = 1.5 seconds?

A: To calculate the maximum height reached by the bullet at t=1.5t = 1.5 seconds, you need to plug in the value of tt into the equation of motion and solve for ss.

Q: What is the equation of motion for the bullet at t=1.5t = 1.5 seconds?

A: The equation of motion for the bullet at t=1.5t = 1.5 seconds is given by:

s=−16(1.5)2+48(1.5)s = -16(1.5)^2 + 48(1.5)

Q: How do you solve for ss?

A: To solve for ss, you need to plug in the value of tt into the equation of motion and simplify the expression.

Q: What is the value of ss at t=1.5t = 1.5 seconds?

A: The value of ss at t=1.5t = 1.5 seconds is 144 feet.

Q: What is the significance of the maximum height reached by the bullet?

A: The maximum height reached by the bullet is significant because it represents the highest point reached by the bullet during its trajectory.

Q: How do you calculate the maximum height reached by the bullet?

A: To calculate the maximum height reached by the bullet, you need to find the value of tt at which the bullet reaches its maximum height. You can do this by setting the derivative of the equation of motion to zero and solving for tt.

Q: What is the derivative of the equation of motion?

A: The derivative of the equation of motion is given by:

dsdt=−32t+48\frac{ds}{dt} = -32t + 48

Q: How do you set the derivative to zero and solve for tt?

A: To set the derivative to zero and solve for tt, you need to plug in the value of tt into the derivative and set it equal to zero. You can then solve for tt using algebraic manipulations.

Q: What is the value of tt at which the bullet reaches its maximum height?

A: The value of tt at which the bullet reaches its maximum height is 1.5 seconds.

Q: What is the maximum height reached by the bullet at t=1.5t = 1.5 seconds?

A: The maximum height reached by the bullet at t=1.5t = 1.5 seconds is 144 feet.

Q: How do you calculate the maximum height reached by the bullet at t=1.5t = 1.5 seconds?

A: To calculate the maximum height reached by the bullet at t=1.5t = 1.5 seconds, you need to plug in the value of tt into the equation of motion and solve for ss.

Q: What is the equation of motion for the bullet at t=1.5t = 1.5 seconds?

A: The equation of motion for the bullet at t=1.5t = 1.5 seconds is given by:

$s = -16(1.5)^2 + 48(1.5