The displacement function of a rock thrown straight up in the air is given by f(t) = -2t2 + 15.7t meters, where t is measured in seconds. Find the total distance travelled by the rock when it reaches the ground.

Step-by-step explanation:

The height of the rock as it  reaches the ground is zero i.e f(t) = 0

Substitute into the function to get t first as shown;

f(t) = -2t^2 + 15.7t

0 = -2t^2 + 15.7t

2t^2 = 15.7t

2t = 15.7

t = 15.7/2

t = 7.85secs

Get the total distance travelled by substituting t = 7.85 into the modeled equation;

Recall that:

f(t) = -2t^2 + 15.7t

f(7.85) = -2(7.85)^2 + 15.7(7.85)

f(7.85) = -123.245+123.245

Hence the total distance travelled by the rock when it reaches the ground is 0m

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2