An arrow is shot straight up from the edge of a cliff with an initial velocity of 50 ft/s. Answer the following questions using English units. The acceleration due to gravity has a magnitude of 32.2 ft/s.
A. Find the time to reach the highest point.
B. Find the displacement at the highest point.
C. Find the height when the velocity is 25 ft/s.
A. t = 50 ft/s / 32.2 ft/s²
B. the velocity is zero at the peak
... so the average ascending velocity is 25 ft/s
... the displacement is 25 ft/s multiplied by the time from part A
C. the time is half of the time from part A
... the average velocity is (50 + 25)/2 ft/s
... the height is velocity times time
Can someone answer this step by step please!!!
A. To find the time to reach the highest point, we need to find the time it takes for the arrow to reach a velocity of 0 ft/s.
Using the formula:
v = u + at
where:
v = final velocity (0 ft/s)
u = initial velocity (50 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2)
0 = 50 + (-32.2)t
Rearranging the equation:
32.2t = 50
Dividing both sides by 32.2:
t ≈ 1.55 seconds
So, it takes approximately 1.55 seconds for the arrow to reach the highest point.
B. The displacement at the highest point can be found using the formula:
s = ut + (1/2)at^2
where:
s = displacement
u = initial velocity (50 ft/s)
t = time to reach the highest point (approximately 1.55 seconds)
a = acceleration due to gravity (-32.2 ft/s^2)
s = 50(1.55) + (1/2)(-32.2)(1.55)^2
s ≈ 77.5 - 38.25
s ≈ 39.25 ft
Therefore, the displacement at the highest point is approximately 39.25 ft.
C. To find the height when the velocity is 25 ft/s, we can use the formula:
v = u + at
where:
v = final velocity (25 ft/s)
u = initial velocity (50 ft/s)
a = acceleration due to gravity (-32.2 ft/s^2)
25 = 50 + (-32.2)t
Rearranging the equation:
32.2t = 50 - 25
32.2t = 25
Dividing both sides by 32.2:
t ≈ 0.778 seconds
Now, with the time, we can find the height using the formula:
s = ut + (1/2)at^2
where:
s = height
u = initial velocity (50 ft/s)
t = time to reach the velocity of 25 ft/s (approximately 0.778 seconds)
a = acceleration due to gravity (-32.2 ft/s^2)
s = 50(0.778) + (1/2)(-32.2)(0.778)^2
s ≈ 38.9 - 10.08
s ≈ 28.82 ft
Therefore, the height when the velocity is 25 ft/s is approximately 28.82 ft.
A. To find the time to reach the highest point, we can use the equation of motion:
v = u + at
where
v = final velocity (0 ft/s at the highest point since the arrow momentarily stops),
u = initial velocity (50 ft/s),
a = acceleration due to gravity (-32.2 ft/s^2, negative since gravity acts opposite to the direction of motion),
t = time.
Rearranging the equation to solve for t:
t = (v - u) / a
Substituting the given values:
t = (0 - 50) / -32.2
t = 1.55 seconds
Therefore, it takes approximately 1.55 seconds for the arrow to reach the highest point.
B. The displacement at the highest point can be determined using the equation:
s = ut + (1/2)at^2
where
s = displacement,
u = initial velocity (50 ft/s),
t = time (1.55 seconds),
a = acceleration due to gravity (-32.2 ft/s^2).
Substituting the known values:
s = (50)(1.55) + (1/2)(-32.2)(1.55)^2
s = 77.5 - 39.8275
s = 37.6725 feet
Hence, the displacement at the highest point is approximately 37.6725 feet.
C. To determine the height when the velocity is 25 ft/s, we can use the equation:
v^2 = u^2 + 2as
where
v = final velocity (25 ft/s),
u = initial velocity (50 ft/s),
a = acceleration due to gravity (-32.2 ft/s^2),
s = displacement (height).
Rearranging the equation to solve for s:
s = (v^2 - u^2) / (2a)
Substituting the known values:
s = (25^2 - 50^2) / (2 * -32.2)
s = (625 - 2500) / (-64.4)
s = -1875 / -64.4
s = 29.09 feet
Therefore, the height when the velocity is 25 ft/s is approximately 29.09 feet.