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? Use the formula: h=−16t²+vt+h0. (Round your answer to the nearest tenth.)

Answer 1

h(t) = -16t2 + 186t + 43

at the ground h = 0

hence; -16t2 + 186t + 43 = 0

solving this quadratic equation using the quadratic formula ; a = -16, b = 186, c = 43 ; x = (-b +-(b2 - 4ac)1/2)/2a

gives t = 11.8 seconds to the nearest tenth (note that the negative root has no practical significance)

Answer 2

time required for arrow to reach ground is 10.8 sec

when t = 10.8 seconds to the nearest tenth

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