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An arrow is shot vertically upward from a platform 19ft high at a rate of 163ft/sec. When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0. (Round your answer to the nearest tenth.)

Respuesta :

Answer: 10.3 s

Step-by-step explanation:

Given

The height of the arrow is given by [tex]h=-16t^2+v_ot+h_o[/tex]

At  [tex]t=0, h=19\ ft[/tex]and [tex]v_o=163\ ft/s[/tex]

When arrow hits ground, h becomes 0 i.e.

[tex]\Rightarrow 0=-16t^2+163t+19\\\Rightarrow 16t^2-163t-19=0\\\\\Rightarrow t=\dfrac{163\pm \sqrt{(-163)^2+4(16)(19)}}{2(16)}\\\\\Rightarrow t=-0.115,10.30\ s[/tex]

Neglecting the negative value of t

So, the arrow hits the ground after 10.30 s