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April shoots an arrow upward at the speed of 80 feet per second from a platform 50 feet high. The pathway of the arrow can be represented by using the equation h(t)=-16t^2+V₀t+h₀

a) What is the maximum height of the arrow?
b) About how many seconds will it take for the arrow it reach the ground?
c) How many seconds after launch will the arrow be 114 feet high?
I need this quickly please

User Tolanj
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a) To find the maximum height of the arrow, we need to find the vertex of the parabolic function. The vertex is located at t = -b/2a, where a = -16, b = V_0 = 80, and h_0 = 50. Therefore, t = -80 / (2 * (-16)) = 2.5 seconds. Plugging t = 2.5 into the equation gives:

h(2.5) = -16(2.5)^2 + 80(2.5) + 50 = 125 feet.

Therefore, the maximum height of the arrow is 125 feet.

b) To find how long it takes for the arrow to reach the ground, we need to find the value of t when h(t) = 0. Substituting h(t) = 0 and solving for t, we get:

0 = -16t^2 + 80t + 50
t^2 - 5t - 50/16 = 0
(t - 10/4)(t + 5/4) = 0

Therefore, t = 2.5 seconds or t = -1.25 seconds. Since the time cannot be negative, the arrow will hit the ground after approximately 2.5 seconds.

c) To find the time when the arrow is 114 feet high, we need to solve the equation h(t) = 114. Substituting h(t) = 114 and simplifying, we get:

-16t^2 + 80t + 50 = 114
-16t^2 + 80t - 64 = 0
t^2 - 5t + 4 = 0
(t - 4)(t - 1) = 0

Therefore, t = 1 second or t = 4 seconds. Since the arrow is at a height of 114 feet after it has been launched, the answer is t = 1 second.
User Lies
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