Question

A ball is tossed in the air in such a way that the path of the ball is modeled by the equation $y\,\, = \, - {x^2}\, + \,6x$ where y represents the height of the ball in feet and x is the time in seconds. At what time x will the ball reach the ground again?
A.6
B.2
C.3
D.4
E.1

Answer 1

Hint: In this question, we have given an equation. First, we will find the highest point where the ball reaches. So, for this we will find the maxima of the function, $\dfrac{{dy}}{{dx}}$= 0.

Complete step-by-step answer:
We have $y\,\, = \, - {x^2}\, + \,6x$
Now, we will check that this is maxima $\dfrac{{{d^2}y}}{{d{x^2}}}$< 0
$\dfrac{{{d^2}y}}{{d{x^2}}}$< 0Here, we will find the highest point where the ball reaches, so we need to find the maxima of the function.
At maxima, $\dfrac{{dy}}{{dx}}$= 0, $\dfrac{{{d^2}y}}{{d{x^2}}}$< 0
We have $y\,\, = \, - {x^2}\, + \,6x$
Differentiating both sides with respect to x,
$\dfrac{{d( - {x^2} + 6x)}}{{dx}}$= 0
$ \Rightarrow $-2x + 6 = 0
$ \Rightarrow $x = 3
$ \Rightarrow \dfrac{d}{{dx}}( - 2x + 6) < 0$
$ \Rightarrow $-2 < 0.
Hence, the ball reaches its maximum point in 3 seconds.
Ball will reach the ground again in 3 + 3 = 6 seconds.
Thus, option A is the correct option.

So, the correct answer is “Option A”.

Note: When a ball reaches maximum height then the slope is zero. If the second derivative is less than zero it means it is a local maximum. When the ball is thrown upward then it forms a parabolic path. Given below is a parabolic path.