Mat 205 Week 2
Focus on Application: Week Two
Student’s Name
MAT/205
Due Date
Instructor’s Name
For the example of a ball being thrown up into the sky and then landing on the ground, we can model a quadratic equation to show the path of the projectile at various points in time (projectile motion). That is to say, each point plotted on the graph (parabola) will be a measurement to this effect:
Suppose a ball is thrown into the sky at a velocity of 64ft/sec from an initial height of 100ft. We would set the quadratic equation as (s)0=-gt^2+v0t+h0 and substitue values for gravity, velocity, and initial height to equal 0=-16t^2+64t+100. If we want to find out after how many seconds the ball will land, we can leave the equation set to zero and solve for t, using the quadratic formula. This will give us a solutions for t = SQRT(41)/2 (approximately 3.2 seconds) or t = -SQRT(41)/2 (approximately -3.2 seconds). Because we are only interested in positive values and negative values would not make any sense in this application, the ball will land after 3.2 seconds have passed, meaning that t = +3.2 seconds when the postion is at 0 or ground level; Position (3.2,0) on the …show more content…
It can give us the projectile’s (ball’s) maximum height value in this equation before it begins to descend. This is simply done by finding the value of h for the x-coordinate of the parabola, and then substituting that value into the equation and solving for k, whereas h is equal to –b/2a in the standard ax^2+bx+c quadratic equation. Therefore, in our case, -64/2(-16)=2 is equal to the x-coordinate of the parabola’s maximum value or vertex. Substituting 2 into -16t^2+64t+100 will give us -16(2)^2+64(2)+100=164, which is the vertex’s y-coordinate. Combining them gives the maximum value of the ball’s flight; the height at 164 feet and the time at 2 seconds, which combine at the vertex point (2,164) on the Cartesian
Student’s Name
MAT/205
Due Date
Instructor’s Name
For the example of a ball being thrown up into the sky and then landing on the ground, we can model a quadratic equation to show the path of the projectile at various points in time (projectile motion). That is to say, each point plotted on the graph (parabola) will be a measurement to this effect:
Suppose a ball is thrown into the sky at a velocity of 64ft/sec from an initial height of 100ft. We would set the quadratic equation as (s)0=-gt^2+v0t+h0 and substitue values for gravity, velocity, and initial height to equal 0=-16t^2+64t+100. If we want to find out after how many seconds the ball will land, we can leave the equation set to zero and solve for t, using the quadratic formula. This will give us a solutions for t = SQRT(41)/2 (approximately 3.2 seconds) or t = -SQRT(41)/2 (approximately -3.2 seconds). Because we are only interested in positive values and negative values would not make any sense in this application, the ball will land after 3.2 seconds have passed, meaning that t = +3.2 seconds when the postion is at 0 or ground level; Position (3.2,0) on the …show more content…
It can give us the projectile’s (ball’s) maximum height value in this equation before it begins to descend. This is simply done by finding the value of h for the x-coordinate of the parabola, and then substituting that value into the equation and solving for k, whereas h is equal to –b/2a in the standard ax^2+bx+c quadratic equation. Therefore, in our case, -64/2(-16)=2 is equal to the x-coordinate of the parabola’s maximum value or vertex. Substituting 2 into -16t^2+64t+100 will give us -16(2)^2+64(2)+100=164, which is the vertex’s y-coordinate. Combining them gives the maximum value of the ball’s flight; the height at 164 feet and the time at 2 seconds, which combine at the vertex point (2,164) on the Cartesian