An arrow is launched upward with a velocity of 224
feet per second from the top of a 90
-foot platform. What is the maximum height attained by the arrow?
height = -16t^2 + 224t + 90
completing the square:
height = -16(t^2 - 14t + ....) + 90
= -16t^2(t^2 - 14t + 49 - 49) + 90
= -16(t-7)^2 + 784+90
= -16(t-7)^2 + 874
so the arrow will reach a max height of 874 ft after 7 seconds
or, using Calculus
d(height)/dt = -32t + 224 = 0 for a max of height
32t = 224
t = 7
sub 7 into height = -16t^2 + 224t + 90
= -16(49) + 224(7( + 90
= 874
time to max ... 224 / 32 ... seconds
average velocity ... 224 / 2
velocity * time = height above platform
remember to add in the platform height
Oh, I can't resist a question about heights and arrows! Well, if the arrow is going up, up, and away, it will start slowing down due to the force of gravity until it reaches its maximum height. So, let's calculate that height, shall we?
To find the maximum height, we need to apply some good old physics. We know that the initial vertical velocity is 224 feet per second, but that will decrease as the arrow goes up. Once the arrow reaches its maximum height, its vertical velocity will be zero. So, we can use the vertical motion equation:
v² = u² + 2as
Where:
v = final velocity (zero at the top)
u = initial velocity (224 feet per second)
a = acceleration due to gravity (32 feet per second squared)
s = displacement or height
Rearranging the equation:
s = (v² - u²) / (2a)
Plugging in the values:
s = (0 - 224²) / (2 * -32)
After some calculations, we find that the maximum height attained by the arrow is approximately 700.5 feet.
So, it seems our arrow reached quite the lofty height! I hope it didn't develop a fear of heights though, because it had quite the drop back down.
To find the maximum height attained by the arrow, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 ft/s at the highest point)
u = initial velocity (224 ft/s)
a = acceleration (due to gravity, -32 ft/s^2 as the arrow moves upward)
s = displacement or change in height
Plugging in the values, the equation becomes:
0^2 = 224^2 + 2(-32)s
Simplifying further:
0 = 50176 - 64s
Rearranging the equation to solve for s:
64s = 50176
s = 50176 / 64
s ≈ 784 ft
Therefore, the maximum height attained by the arrow is approximately 784 feet.
To find the maximum height attained by the arrow, we need to use the principles of projectile motion. We can start by analyzing the motion of the arrow in two separate parts: the vertical motion and the horizontal motion.
In the vertical motion, the arrow goes up and then falls back down due to gravity. At the maximum height, the vertical velocity will be zero, and the arrow will start coming back down.
We can use the equation of motion to find the maximum height:
v_f^2 = v_i^2 + 2ad
Where:
v_f is the final velocity (which is zero at the maximum height),
v_i is the initial vertical velocity (224 feet per second in this case),
a is the acceleration due to gravity (-32 feet per second squared),
and d is the vertical displacement (which we want to find).
Rearranging the equation, we have:
0 = (224 feet per second)^2 + 2(-32 feet per second squared)d
Simplifying, we get:
0 = 50176 feet^2 per second^2 - 64 feet per second squared * d
Solving for d, we have:
d = 50176 feet^2 per second^2 / 64 feet per second squared
d = 784 feet
Therefore, the maximum height attained by the arrow is 784 feet.