An arrow is shot vertically upward from a platform 39ft high at a rate of 153ft/sec .

Question

(Round your answer to the nearest tenth.)

When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0 .

Answer 1

They give you an equation with many variables. First thing is to look at what we know, what we don't know, and what we want to know within the equation.

1. h₀ is the initial height. The problem states that you are one a platform 39ft up in the air, so h₀ = 39
2. v₀ is the initial velocity. The problem directly gives it to you as 153 ft/sec, so v₀ = 153
3. t is the time. The question is asking for when the arrow hits the ground. This is what we want to find.
4. h is the height. We want to know when the arrow hits the ground. The ground would be a height of 0.

0 = -16t² + 153t + 39

This is in fact a quadratic equation. You can use your preferred method to solve it. I don't believe this is factorable so you might need to put it in a graphing calculator or use the quadratic equation. at the end you should get two answers for time, but only one is positive. Something around t = 9.8

Answer 2

h(t) = 0 = -16t^2 +153t +39

16t^2 -153t -39 = 0

t = 153/32 + (1/32)sqr(153^2 +64(39))

t = (153 + 161.1)/32

t= about 314.1/32 = 9.815625 =

about 9.8 seconds to hit the ground