An arrow is shot vertically upward at a rate of 220ft/s. use the projectile formula h=? 16t2+v0t to determine at what time(s), in seconds, the arrow is at a height of 400ft. round your answer(s) to the nearest tenth of a second.

t1 =  2.2 s

t2 = 11.5 s

Step-by-step explanation:

The general quadratic formula for these cases is

h(t) = h0 + v*t + 0.5*a*t^2

Where

a = the acceleration of the body. In this case, since the arrow is going upwards, we take the acceleration of the gravity.

a = -32.2 feet/s^2 = -9.8 m/s^2

v = the velocity

v = 220 ft/s

h0 =  the initial distance. We assume it launched from ground

h0 = 0

h(t) =  - 16.1*t^2 + 220*t

See, attached picture

When the height h = 400 ft

- 16.1*t^2 + 220*t = 400

t1 = 2.159 s ≈ 2.2 s

t2 = 11.505 s ≈ 11.5 s

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)

probably the last option

step-by-step explanation: