An arrow is shot vertically upward from a platform 21ft
high at a rate of 228ft/sec
. When will the arrow hit the ground? Use the formula: h=−16t2+v0t+h0
. (Round your answer to the nearest tenth.)
We can use the formula h = -16t^2 + v0t + h0 where:
h = height of the arrow (which will be 0 when it hits the ground)
t = time taken for the arrow to hit the ground
v0 = initial velocity of the arrow (228 ft/sec)
h0 = initial height of the arrow (21 ft)
Plugging in the given values, we have:
0 = -16t^2 + 228t + 21
This is a quadratic equation that we can solve for t. Rearranging the equation, we get:
16t^2 - 228t - 21 = 0
We can solve this quadratic equation by factoring or by using the quadratic formula. Since factoring may not be possible, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 16, b = -228, and c = -21. Plugging in these values, we get:
t = (-(-228) ± √((-228)^2 - 4(16)(-21))) / (2(16))
t = (228 ± √(52044 + 1344)) / 32
t = (228 ± √(53388)) / 32
To find the time when the arrow hits the ground, we need to take the positive value:
t = (228 + √(53388)) / 32
t ≈ 3.6 seconds
Therefore, the arrow will hit the ground approximately 3.6 seconds after being shot vertically upward.