An arrow is shot vertically upward from a platform 33ft high at a rate of 214ft/sec. when will the arrow hit the ground? use the formula: h=−16t2+v0t+h0. (round your answer to the nearest tenth.)

Time = 11.1 seconds

Explanation:

h = -16t^2 + v(0)t + h(0)

but we are given;

v(0) = 174 ft/sec

h(0) = 33 ft

Find h = 0:

Therefore;

0 = -16t^2 + 174t + 33

16t^2 - 174t - 33 = 0

Use the quadratic equation:

a = 16

b = -174

c = -33

t = (174 +/- sqrt((-174)^2 - 4(16)(-33)))/2(16)

t = (174 +/- sqrt(30276 + 2112))/32

t = (174 +/- sqrt(32388))/32

t = (174 +/- 179.97)/32

t = 353.97/32 or -5.97/32

t = 11.1 or -.2

Therefore;

The arrow will hit the ground at:  

t = 11.1 sec

heats surface water up

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures