An arrow is shot vertically up into the air with an initial vertical velocity of 60 m/s, and its height is given by h=-5t^2+60t where h is in meters and t is in seconds. How high does the arrow go? How long does the arrow stay in flight?

Step-by-step explanation:

Velocity is the change in displacement of an object.

Velocity = dh/dt

Given the equation modelled by the height travelled by the area as h=-5t^2+60t

v = dh/dt = -10t + 60

At the maximum height, the velocity is zero i.e. -10t+60 =0

-10t = -60

t = 6secs

Substituting t = 6 secs into the modeled equation to know how high the arrow goes;

h=-5t^2+60t

h = -5(6)²+60(6)

h = -180+360

h = 180m

Hence, the arrow travels 180m high.

To determine how long that the arrow stays in flight, we will apply the formula:

H = ut+1/2gt²

180 = 60t+1/2(9.8)t²

180 = 60t +4.9t²

49t²+600t -1800 = 0

t = -600±√600²-4(49)(-1800)/98

t = -600±√360000+352800/98

t = -600±√712800/98

t = -600+844.28/98

t = 244.28/98

t = 2.49secs

Hence, the arrow stays 2.49secs in the flight.

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step-by-step explanation:

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)