Question

The path when a stone is thrown can be modelled by = −16 2 + 10 + 4, where (in feet) is the height of the stone seconds after it is released.
a. Graph the function.


b. Determine the maximum height reached by the stone.


c. How long will it take the stone to reach its maximum height?​

Answer 1

Assuming that

f(t) = -16t² + 10t + 4

f(t) is the height (in feet) attained by the ball after t seconds of it's launch into air.

a) To graph the equation -16t² + 10t + 4 , first we need to find the roots

-16t² + 10t + 4 = 0

-8t² + 5t + 2 = 0

t = [ -5 + √25 - 4(-8)(2) ] / -16 & t = [ -5 - √25 - 4(-8)(2) ] / -16

t = 5/16 - √89 / 16 & t = 5/16 + √89 / 16

Since , -16t² + 10t +4 is a quadratic polynomial of the form -ax² + bx + c

Therefore the graph of this polynomial is a upside down parabola where the zeroes are (5/16-√89/16 , 0) & (5/16+√89/16 , 0)

See the attachment for the graph of f(t) = -16t² +10t +4

b) The maximum height attained by the stone is equal to the f(t) value at maxima of the curve f(t) = -16t² + 10t + 4

Differentiating the function f(t) w.r.t 't'

f'(t) = ∂(f(t)) / ∂t

f'(t) = -32t + 10 ∀ t ∈ ℝ

Solving the equation f'(t) = 0 for t , we get the values of t where the gradient function f'(t) is equal to 0 i.e. slope of f(t) = 0

f'(t) = -32t + 10 = 0

t = 5/16

Differentiating the gradient function f'(t) w.r.t 't'

f''(t) = ∂(f'(t))/∂t

f''(t) = -32 ∀ t ∈ ℝ

f''(5/16) = -32

Therefore there is a global maxima at t = 5/16 on the curve

f(t) = -32t² + 10t + 4

Thus the maximum height reached

by the stone = f(5/16) = -( 25 / 16 ) + 10 ( 5 /16) + 4 = 5.5625 ft

c) The time taken by the ball to reach the maximum height is

equal 5/16 or 0.3125 seconds

Answer 2

Question: The path when a stone is thrown can be modelled by $y=-16x^{2}+10x+4,$, where (in feet) is the height of the stone seconds after it is released.

      a. Graph the function.

       b. Determine the maximum height reached by the stone.

       c. How long will it take the stone to reach its maximum height?

Concept: We can use the calculus method to find the maximum height and time,

Given that: A stone is thrown can be modelled by $y=-16x^{2}+10x+4,$    $y$ is the height, time is $x$ second released.

Solution:

a. Graph the function: Given below in image end of the answer,

b. Determine the maximum height reached by the stone:

Given that, $y=-16x^{2}+10x+4,$

Differentiate with respect to $x$ both side

                            $\frac{dy}{dx}=(\frac{-16x^{2}+10x+4 }{dx})\\\frac{dy}{dx}=d(\frac{-16x^{2} }{dx})+d(\frac{10x}{dx})+d(\frac{4}{dx})\\\\\frac{dy}{dx}= -32x+10$

hence,   $\frac{dy}{dx}=0$

then,

                             $-32x^{}+10=0\\ -32x=-10\\ 32x=10\\ x=\frac{10}{32}$

or we can write, $x= \frac{5}{16}$

hence, stone the maximum height reached by $x=\frac{5}{16}$

c. How long will it take the stone to reach its maximum height:

given that $y=-16x^{2}+10x+4$

                        $\frac{dy}{dx}=-32x+10$   from above question

 set the derivative equal to zero then,

                             $-32x+10=0\\-32x=-10\\32x=10\\x=\frac{10}{30}\\x=0.3125$

hence it will take $0.3125$ seconds for the stone reach to maximum height.

Final answer: Graph mention end of the answer, time taking for maximum height is $0.3125$seconds and maximum height when $x=\frac{5}{16}$feet,

#SPJ3

                                   Fig: Graph of the function