If an archer shoots an arrow straight upward with an initial velocity of 160

ft/sec from a height of 8 feet, its height above the ground in feet at time t in seconds is

given by the function h(t)= -16t^2+160t +8. what is the maximum height reached by the arrow

maximum height of the arrow will be 408 m

Step-by-step explanation:

We have given the function of height as

We know that first derivative of height will be velocity

So first derivative of height

We know that at maximum height velocity will be zero

t = 5 sec

Sp arrow will be at maximum height at t = 5 sec

So height at t = 5 sec will be

So maximum height of the arrow will be 408 m

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3