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?
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(
To find the maximum height attained by the arrow, we can use the principles of projectile motion. The arrow goes up against gravity until its velocity becomes zero at the highest point, and then it starts to fall back down.
First, we need to find the time it takes for the arrow to reach its maximum height. We can use the equation:
v = u + at
where
v = final velocity (which is 0 at the highest point)
u = initial velocity (54 feet per second)
a = acceleration due to gravity (-32 feet per second squared, considering it is directed downwards)
Rearranging the equation to solve for time (t):
t = (v - u) / a
t = (0 - 54) / (-32)
t ≈ 1.6875 seconds
Now, we can find the maximum height by using the equation:
s = ut + 0.5at^2
where
s = displacement (maximum height)
u = initial velocity (54 feet per second)
t = time taken (1.6875 seconds)
a = acceleration due to gravity (-32 feet per second squared)
Plugging in the values, we get:
s = (54 * 1.6875) + 0.5 * (-32) * (1.6875)^2
s ≈ 45.5625 + (-27) * 2.840390625
s ≈ 45.5625 - 76.685250000000
s ≈ -31.122750000000
Since we are finding the height, the negative value doesn't make sense in this context, so we ignore the negative sign.
Therefore, the maximum height attained by the arrow is approximately 31.122 feet.