In Week 4 we learned about quadratic equations. In physics a quadratic equation can be used...

In Week 4 we learned about quadratic equations. In physics a quadratic equation can be used to model projectile motion. Projectile motion can describe the movement of a baseball after it has been hit by a bat, or the movement of a cannonball after it has been shot from a cannon. A penny falling from the Empire State Building can even be modeled with this equation!

The projectile motion equation is s(t)=-16t^2+vt+h where s(t) represents the distance or height of an object at time t, v represents the initial speed of the object in ft/s, and h is the initial height of the object, measured in feet.

If an object is starting at rest, then v=0 (such as for a penny being dropped from a building). If the object is starting from the ground, h=0. The baseball or cannonball situations, each have an initial velocity. For example, the initial velocity of the baseball is based on the speed at which the ball comes at you (the speed of the pitch).

Come up with a situation that you can model with this equation. Describe the situation, and tell us what v and h are. Fill in the values so that you have a quadratic equation. If you do research to find initial velocities, include the links to the websites where you found that information. If you would like to make up your own numbers as well, you can (be creative)!

Once you have your equation, find the maximum height as well as the time it takes to reach that maximum. Then use your equation to find when the object hits the ground (i.e. the x-intercepts).

Finally, use those three points as well as the initial height to sketch a graph. You can take a photo of it and include the image, or use an online graphing calculator and take a screenshot if that is easier.

Solutions

Expert Solution

A cannon ball is fired into the air from the top of a fort. The height (h) of the cannon ball above the ground in feet , t seconds after the cannon is fired, is given by the function h(t)= ?t² + 300t + 10000. Here, 10000 feet is the height of the fort and the initial velocity of the cannon ball is 300 ft./sec.

The path of the cannon ball is given by h(t)= ?t² + 300t + 10000 which is the standard form of the equation of a parabola which opens downwards. In vertex form, we have h(t) = -(t² -2*150t +22500)+10000+22500 = -(t-150)² +32500. This is the equation of a parabola which opens downwards , with vertex at (150,32500). Since the vertex is the highest point of a parabola which opens downwards, the maximum height achieved by the cannon ball is 32500 ft. The time taken to achieve this height is 150 seconds or, 2.5 minutes ( the t- coordinate of the vertex). Further, if the cannon ball hits the ground after t second, then ?t² + 300t + 10000 = 0 or, t² -300t – 10000 = 0 so that t = [300{(-300)² -4*1*(-10000)]/2*1 = = [300

A graph of the trajectory of the cannon ball is attached.

milcah answered 1 year ago

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