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

the displacement produced by the body.  

the amount of force subjected to the body.  

the angle between the direction of force and displacement

spare some change and give it to the poor

explanation:

plz : ((