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

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?

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2 weeks 2021-09-10T10:16:49+00:00 1 Answer 0

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

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

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