Question

You and a friend each kick a soccer ball up into the air. The polynomials -16 T squared +6040+2 and -16 T squared +5060+1 represent the height of the soccer ball after T seconds

Answer 1

Answer:

You and a friend each kick a soccer ball up into the air. The polynomials -16×T²+6040×T+2 and -16×T²+5060×T+1 represent the height of the soccer ball after T seconds

a. How long does it take for the ball to attain maximum height

b. Find the maximum height reached by each kick of the ball

c. What is the difference in height between the two polynomials (between the highest heights of the two ball kicks)

The answers to the question as follows

a. First kick = 188.75 s, second kick = 158.125 s

b. First kick = 570027 meters, second kick = 400057.25 meters

c. 169969.75 meters

Step-by-step explanation:

a. To solve the question, we write out the equations thus

The polynomials are

For the first player we have

Height of soccer ball = -16×T²+6040×T+2 and for the second player we have

Height of soccer ball = -16×T² + 5060×T+1

To find out which ball achieves the highest height we have to look for the point at which the slope of the curve described by the parabola is equal to zero

Therefore since slope = $\frac{y_{2} -y_{1} }{x_{2} - x_{1} }$ that is

(Change in the y coordinates)/(equivalent change in x coordinates)

Hence at the top of the curve, just as it starts the downwards movement, we have a very small point at which y₂ - y₁ = 0 and at this point

$\frac{y_{2} -y_{1} }{x_{2} - x_{1} }$ =$\frac{0}{x_{2} - x_{1} }$= 0

Therefore $\frac{y_{2} -y_{1} }{x_{2} - x_{1} }$ = Δy/Δx or $\frac{dy}{dx}$ = 0

since we are working with T instead of x we put the equation in the following way

y = -16×T²+6040×T+2 therefore $\frac{dy}{dT}$ =$\frac{d(-16T^{2} +6040T+2 )}{dT}$ following the rules of differentiation this gives

-32×T+6040 for which as noted =0, hence the time to reach the highest point  → -32×T+6040 = 0 or T = 188.75 s

Similarly for the second person that played the ball we have

$\frac{dy}{dT}$ = $\frac{d(-16T^{2} +5060T+1 )}{dT}$ = -32×T+5060= 0 or T = 158.125 s

b. The maximum height reached of the ball by the first kicker= -16×(188.75)²+6040×188.75+2 = 570027 meters

Similarly for the second person that played the ball we have

$\frac{dy}{dT}$ = $\frac{d(-16T^{2} +5060T+1 )}{dT}$ = -32×T+5060= 0 or T = 158.125 s

and maximum height reached = -16×T² + 5060×T+1 = -16×(158.125)² + 5060×(158.125)+1 = 400057.25 meters

c. Therefore the ball goes highest for the first kick of the ball than for the second by

570027 - 400057.25 = 169969.75 meters