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A.) y = ut - 1/2gt^2 + 12
y = 6t - 1/2(9.8)t^2 + 12
y = 6t - 4.9t^2 + 12
b.) The y-intercept is 12
c.) The y-intercept represent the height of the ball at time t = 0
d.) At the time the ball reaches the ground, y = 0
-4.9t^2 + 6t + 12 = 0
t = 2.29 seconds
Therefore, it took the ball 2.29 seconds to reach the ground.
Answer:
125feet
Step-by-step explanation:
Given the equation that modeled the height expressed as h = -16t^2 + 80t + 25, where h is the height and t is the time in seconds.
The arrow reaches the maximum height at dh/dt = 0
dh/dt = -32t + 80
0= -32t+80
32t = 80
t = 80/32
t = 2.5secs
substitute t = 2.5 into the formula;
h = -16t^2 + 80t + 25
h = -16(2.5)^2 + 80(2.5) + 25
h = -16(6.25)+225
h = -100+225
h = 125
Hence the maximum height the arrow reaches is 125feet
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