Answer :
You didn't include the graph of Leslis's basball. So you will have to make the final comparison.
To find the maximum height of Devin's baseball you use your knweledge of quadratic functions.
h(t) = - 16t^2 + 79t + 5.
The maximum height is at the vertex of the parabole.
The vertex is at a t-coordinate in the mid value of the roots of the equation.
To find the roots you can either factor the function of use the quadratic equation.
The quadratic equation is [-b +/+ √(b^2 - 4ac)]/2a
where a = -16, b = 79 and c = 5.
The result is t = -1/16 and t = 5
The mid value is t= [-1/16 + 5]/2 = 79/32 ≈ 2.47
And the correspondig height is h(79/32) = -16[79/32]^2 + 79[79/32]+5 ≈ 102.5
Now compare 102.5 with the highest value on the graph and there you have who is right.
The time that Devin's baseball was in the air corresponds to the time of the second root (when the ball comesback to h(t) = 0). That is t = 5 seconds.
To find the maximum height of Devin's baseball you use your knweledge of quadratic functions.
h(t) = - 16t^2 + 79t + 5.
The maximum height is at the vertex of the parabole.
The vertex is at a t-coordinate in the mid value of the roots of the equation.
To find the roots you can either factor the function of use the quadratic equation.
The quadratic equation is [-b +/+ √(b^2 - 4ac)]/2a
where a = -16, b = 79 and c = 5.
The result is t = -1/16 and t = 5
The mid value is t= [-1/16 + 5]/2 = 79/32 ≈ 2.47
And the correspondig height is h(79/32) = -16[79/32]^2 + 79[79/32]+5 ≈ 102.5
Now compare 102.5 with the highest value on the graph and there you have who is right.
The time that Devin's baseball was in the air corresponds to the time of the second root (when the ball comesback to h(t) = 0). That is t = 5 seconds.