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## 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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## Answers ( )

Answer:time required for arrow to reach ground is 10.8 secwhen t = 10.8 seconds to the nearest tenthStep-by-step explanation:Given values areVelocity = 190 ft/sec height = 16 ftGiven formula ish=−16t²+vt+h0adding the values, we geth(t)= -16t²+190t+16so we have to find when the hit he ground, so at ground the height will be 00= -16t²+190t+16-16t²² + 190t + 16= 0using the quadratic formulax = (-b +(b2 – 4ac)1/2)/2avaluesa = -16, b = 190, c = 16x= -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 significanceand we have t = 10.8 seconds to the nearest tenth