An arrow is shot vertically upward from a platform 8ft
high at a rate of 157ft/sec
. When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0
. (Round your answer to the nearest tenth.)
To find when the arrow will hit the ground, we need to find the time when the height is equal to 0.
Using the formula h = -16t^2 + v0t + h0, where h is the height, t is the time, v0 is the initial velocity, and h0 is the initial height, we can substitute the given values:
0 = -16t^2 + 157t + 8
Now, we can solve this quadratic equation to find the value of t. We can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this equation, a is the coefficient of t^2 (-16), b is the coefficient of t (157), and c is the constant term (8).
Plugging in the values, we get:
t = (-157 ± √(157^2 - 4(-16)(8)))/(2(-16))
Simplifying the equation further:
t = (-157 ± √(24649 + 512))/(2(-16))
t ≈ (-157 ± √(25161))/(2(-16))
t ≈ (-157 ± 158.7)/(2(-16))
Now, we can calculate the two solutions:
t ≈ (-157 + 158.7)/(-32)
t ≈ 1.7/(-32) ≈ -0.0531
t ≈ (-157 - 158.7)/(-32)
t ≈ -315.7/(-32) ≈ 9.87
Since time cannot be negative in this context, we discard the negative solution.
Therefore, the arrow will hit the ground approximately 9.9 seconds after it was shot.