Question

An arrow is shot vertically upward from a platform 16ft high at a rate of 190ft/sec. When will the arrow hit the ground?

Answer 1

Answer:

time required for arrow  to reach ground is 10.8  sec

when  t = 10.8 seconds to the nearest tenth

Step-by-step explanation:

Given values are  

Velocity = 190 ft/sec                     height = 16 ft

Given formula is  

h=−16t²+vt+h0

adding the values, we get

h(t)= -16t²+190t+16

so we have to find when the hit he ground, so at ground the height will be 0  

0= -16t²+190t+16

-16t²² + 190t + 16= 0  

using the quadratic formula

x = (-b +(b2 - 4ac)1/2)/2a

values  

a = -16, b = 190, c = 16

x= -190 + ((-190)²- 4 (-16)(16) ½ / 2(-16)

x= -190 + ((-190) ²+1024) ½ / 2(-16)

x= -190+ (-190²+ 32²) ½ /-32

taking under root  

x= -190  + ( -190+32)/ -32

x= 190 + (-158)/-32

we have   2 options,

x= 190 + (-158)/-32                       x= 190 - (-158)/-32

x= 190 -158)/-32                         x= 190 +158)/-32

x= -1                                           x= 10.875  

or t = 10.875

so the negative root has no practical significance  

and we have t = 10.8 seconds to the nearest tenth