April shoots an arrow upward at a speed of 80 ft/sec from a platform of 25 ft. High. The pathway of the arrow can be represented
by the equation h = -16t^2 + 80t + 25, where h is the height and t is the time in seconds. What is the maximum height that the arrow reaches
1 answer:
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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