Solving the Trajectory of a Projected Object: A Mathematical Analysis

Objective: This article aims to analyze and solve the trajectory of an object shot straight upward with an initial velocity of 800 feet per second. The primary focus is to determine the time at which the object reaches a specific height of 10000 feet using mathematical methods and the quadratic formula.

Introduction to Projectile Motion

Projectile motion is a fascinating area of physics that deals with the motion of an object under the influence of gravity. When an object is projected vertically upward from a given position, its height changes with time due to the constant downward acceleration due to gravity.

Formulating the Height Equation

The height of an object undergoing constant acceleration a at time t can be described by the equation:

(displaystyle y_t frac{1}{2}at^2 v_0t y_0)

yt is the height of the object at time t. a is the acceleration (in this case, the acceleration due to gravity, -32 feet per second squared). v0 is the initial velocity (800 feet per second). y0 is the initial height (0 feet in this example).

Substituting the given values, the height equation becomes:

(displaystyle y_t frac{-32}{2}t^2 800t)

Simplifying, we get:

(displaystyle y_t -16t^2 800t)

Determining the Time to Reach 10000 Feet

To find the time at which the object reaches a height of 10000 feet, we set yt equal to 10000 feet:

(displaystyle -16t^2 800t 10000)

Rearranging the equation:

(displaystyle -16t^2 800t - 10000 0)

To simplify, we divide the entire equation by -16:

(displaystyle t^2 - 50t 625 0)

Solving the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

(displaystyle t frac{-b pm sqrt{b^2 - 4ac}}{2a})

where a 1, b -50, and c 625. Plugging in these values:

(displaystyle t frac{50 pm sqrt{50^2 - 4 cdot 1 cdot 625}}{2 cdot 1})

(displaystyle t frac{50 pm sqrt{2500 - 2500}}{2})

(displaystyle t frac{50 pm 0}{2})

(displaystyle t frac{50}{2} 25)

Thus, the object will reach a height of 10000 feet at t 25 seconds.

Additional Calculations

For a more detailed analysis, we can also calculate the maximum height:

(displaystyle V_f 0), the final velocity at the peak, and the maximum height using the formula:

(displaystyle text{Max Height} frac{V_0^2}{2g} frac{800^2}{64} 10000 text{ feet})

This confirms that the object can reach a maximum height of 10000 feet.

The height of the object at t 25 seconds is:

(displaystyle y_{25} -16(25)^2 800(25) -10000 20000 10000 text{ feet})

Conclusion

Using the quadratic formula, we determined that the object will reach a height of 10000 feet at 25 seconds. This method can be applied to solve similar problems involving projectile motion and height calculations.

FAQ

What is the height equation for a projectile? (displaystyle y_t frac{1}{2}at^2 v_0t y_0) How does gravity affect the trajectory of an object? (displaystyle text{The acceleration due to gravity is } -32 text{ feet per second squared}). Can you explain the quadratic formula? (displaystyle t frac{-b pm sqrt{b^2 - 4ac}}{2a})

Related Keywords

height equation quadratic formula projectile motion