Asked by Randi
An arrow is launched upward with a velocity of 54 feet per second from the top of a 75 foot building. What is the maximum height attained by the arrow in feet?
Answers
Answered by
Reiny
from your information, the height h is
h = -16t^2 + 54t +75 , where t is in seconds and h is in feet
dh/dt = -32t + 54
= 0 for a max h
32t = 54
t = 54/32 = 27/16 ---> time it took to reach max height = -16(27/16)^2 + 54(27/16) + 75
= 120.56 ft
h = -16(
h = -16t^2 + 54t +75 , where t is in seconds and h is in feet
dh/dt = -32t + 54
= 0 for a max h
32t = 54
t = 54/32 = 27/16 ---> time it took to reach max height = -16(27/16)^2 + 54(27/16) + 75
= 120.56 ft
h = -16(
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