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Question

a baseball player throws a ball from the outfield toward home plate. The ball's height above the ground is modeled by the equation y=-16t^2+48t+6, where y represents height, in feet, and t represents time, in seconds. the ball is initially thrown from a height of 6 feet. What is the maximum height that the ball reaches?

Answer 1

1. First, notice that the equation is that of a parabola.  Projectile motion (such as throwing a ball into the air) follows a parabolic path
2. The question is essentially asking you to find the peak point of the parabola, known as the vertex
3. The formula for the vertex is -b/2a
4. In the equation y = -16t² + 48t + 6, the value of a is -16 and the value of b is 48
5. Plug those values into the equation:
1. vertex = -48/(2*-16) = -48/-32 = 1.5
6. This value of 3 represents the time at which the ball will be at its max height.  To find the height y, plug the value of t =3 into the equation for the parabola:
1. y = -16(1.5)² + 48(1.5) + 6
2. y = -36 + 72 + 6
3. y = 42
7. The maximum height the ball reaches is 42 feet.

 

A second way to solve the problem (especially if you are taking a calculus class) is to use calculus and use the first derivative:

 

1. y = -16t² +48t + 6
2. Take the first derivative and set it equal to zero
1. y' = -32t + 48 = 0
2. -32t = -48
3. t = 1.5
3. Now if you want to be sure that this critical point represents a maximum height (as opposed to a minimum height), you use the second derivative test:
1. y'' = -32
2. Since this is negative, the first derivative test value of 1.5 represents a maximum
4. Again, you have t = 1.5 so to find y, you just plug in the value of t into the original equation.