A helicopter is traveling with a velocity of 12 m/s directly upward. Directly below the helicopter is a very large and very soft pillow. As it turns out, this is a good thing, because the helicopter is lifting a large man. When the man is 20 m above the pillow, he lets go of the rope.
a. What is the speed of the man just before he lands on the pillow?
b. How long is he in the air after he lets go?
c. What is the greatest height reached by the man above the ground? (HINT: this should be greater than 20 m. Why?)
d. What is the distance between the helicopter and the man three seconds after he lets go of the rope?
There are two motions of the man:
1. upward motion with the initial velocity of 12 m/s
h=vₒ²/2g =144 /2•9.8 =7.34 m
t=vₒ/g=12/9.8=1.22 s.
H=20+7.34 =27.34 m
2. Dowmward motion (free fall)
t=sqrt(2H/g)=sqrt(2•27.34/9.8) =2.36 s.
(a) mgH=mv²/2
v=sqrt(2gH) = sqrt(2•9.8•27.34) = 23.15 m/s.
(b) total t = 1.22+2.36=3.58 s
(c) = H = 27.34 m.
(d) the helicopter ia at height
H1= h+v•t= 20+12•3 = 56 m.
The man:
he is falling during
3-1.22 =1.78 s
and covered distance
h1=gt²/2 = 9.8•1.78²/2 =15.53 m
Since he started from the height H=27.34 m, he is at the height
27.34-15.53 =11.81 m
Δ h =56 – 11.81 = 44.19 m
a. The speed of the man just before he lands on the pillow is 12 m/s. Because gravity pulls things downward, the man will continue to accelerate downward at a rate of 9.8 m/s^2 until he reaches the pillow. However, his initial upward velocity of 12 m/s will gradually decrease until it is zero, at which point he will start moving strictly downward with a speed equal to the magnitude of his initial velocity.
b. To find out how long the man is in the air after he lets go, we need to calculate the time it takes for him to reach the pillow. Using the equation s = ut + (1/2)at^2, where s is the distance (20 m), u is the initial velocity (12 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time, we can rearrange the equation to get a quadratic equation: (-4.9)t^2 + 12t - 20 = 0. Solving this equation, we find the positive root to be approximately t = 2.6 seconds.
c. The greatest height reached by the man above the ground will occur at the peak of his trajectory. This happens when his vertical velocity is zero. Using the equation v = u + at, where v is the final velocity (0 m/s), u is the initial velocity (12 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time, we can find the time it takes for the man to reach his peak height. Solving for t, we get t ≈ 1.2 seconds. Plugging this back into the equation s = ut + (1/2)at^2, we find the greatest height is approximately 7.3 meters. This is greater than 20 meters because the man still has some upward velocity when he lets go of the rope, allowing him to continue gaining height before gravity brings him down.
d. Since the man is no longer connected to the helicopter, his motion is solely determined by the force of gravity. We can use the equation s = ut + (1/2)at^2 to find the distance between the helicopter and the man three seconds after he lets go. Plugging in t = 3 seconds and u = 12 m/s, we find the distance to be approximately 11.1 meters. So, three seconds after the man lets go, he is approximately 11.1 meters away from the helicopter.
To solve this problem, we can use the equations of motion. Let's assume that the positive direction is upward.
a. What is the speed of the man just before he lands on the pillow?
To calculate the speed of the man just before he lands on the pillow, we need to find the final velocity (vf) when the man reaches the pillow.
Using the equation of motion:
vf^2 = vi^2 + 2ad
where
vf = final velocity
vi = initial velocity
a = acceleration
d = distance
In this case, the final velocity (vf) will be zero since the man comes to rest after landing.
So, the equation becomes:
0 = vi^2 + 2ad
Rearranging the equation to solve for vi:
vi^2 = -2ad
vi = √(-2ad)
Given:
vi = 12 m/s (upward)
d = 20 m (upward)
a = -9.8 m/s^2 (acceleration due to gravity, pointing downward)
Plugging in the values:
vi = √(-2 * 9.8 m/s^2 * 20 m)
vi = √(-392 m^2/s^2)
vi = -19.8 m/s
Since velocity is a scalar quantity (only magnitude, no direction), the speed will be positive. Therefore, the speed of the man just before he lands on the pillow is 19.8 m/s.
b. How long is he in the air after he lets go?
To calculate the time the man is in the air after he lets go, we can use the equation of motion:
vf = vi + at
where
vf = final velocity (0 m/s)
vi = initial velocity (12 m/s)
a = acceleration (-9.8 m/s^2)
We can solve for time (t):
0 = 12 m/s + (-9.8 m/s^2) * t
Rearranging the equation to solve for t:
9.8 m/s^2 * t = 12 m/s
t = 12 m/s / 9.8 m/s^2
t ≈ 1.22 s
Therefore, the man is in the air for approximately 1.22 seconds after he lets go of the rope.
c. What is the greatest height reached by the man above the ground?
To calculate the greatest height reached by the man above the ground, we can use the equation of motion:
vf^2 = vi^2 + 2ad
Since the final velocity (vf) is zero at the highest point, the equation becomes:
0 = vi^2 + 2ad
Rearranging the equation to solve for d:
d = -vi^2 / (2a)
Given:
vi = 12 m/s
a = -9.8 m/s^2 (acceleration due to gravity)
Plugging in the values:
d = -(12 m/s)^2 / (2 * -9.8 m/s^2)
d = -144 m^2/s^2 / (-19.6 m/s^2)
d ≈ 7.35 m
Therefore, the greatest height reached by the man above the ground is approximately 7.35 meters.
(HINT: The height is positive because the velocity is initially upward and decreases until it reaches zero at the highest point.)
d. What is the distance between the helicopter and the man three seconds after he lets go of the rope?
To calculate the distance between the helicopter and the man three seconds after he lets go of the rope, we need to find the horizontal distance traveled by the man during this time.
Since the man is no longer connected to the helicopter, the only force acting on him is gravity, which causes him to fall vertically. This means that the horizontal velocity remains constant.
Let's assume the horizontal velocity is v_h = 0 m/s (as the helicopter is not moving horizontally).
Using the equation:
d = v_h * t
where
d = horizontal distance
v_h = horizontal velocity (constant)
t = time
We know that t = 3 s.
Plugging in the values:
d = 0 m/s * 3 s
d = 0 m
Therefore, the distance between the helicopter and the man three seconds after he lets go of the rope is 0 meters.
To answer these questions, we need to consider the motion of the man and the helicopter. Let's break it down step-by-step:
a. To determine the speed of the man just before he lands on the pillow, we need to find the velocity at that moment. Since the helicopter is traveling upward with a velocity of 12 m/s, its velocity will contribute to the man's velocity when he lets go. The man's velocity just before he lands can be calculated by subtracting the velocity of the helicopter from his own initial velocity.
To find the man's initial velocity, we can use the equation of motion:
v = u + at
Where:
v = final velocity (unknown)
u = initial velocity (given as 12 m/s)
a = acceleration (acceleration due to gravity, approximately -9.8 m/s^2, considering downward direction)
t = time (unknown)
Since the motion is vertical, the acceleration due to gravity acts in the opposite direction (downward), so it is taken as negative. We can assume the positive direction is upward, so we'll use a negative sign for acceleration.
When the man lets go of the rope, he experiences free fall. We can use the equation for displacement:
s = ut + (1/2)at^2
To find s (displacement), we consider the final height above the pillow (s = -20 m) and solve for t.
Since the man and the helicopter are at the same height when he lets go, we can set s = 0. Now the equation becomes:
0 = 12t + (1/2)(-9.8)t^2
Simplifying the equation:
4.9t^2 - 12t = 0
Factoring out t:
t(4.9t - 12) = 0
This gives us two possibilities:
t = 0 (not relevant in this case)
t = 12/4.9 ≈ 2.45 seconds
Therefore, the man is in the air for approximately 2.45 seconds after he lets go of the rope.
To find the speed of the man just before he lands, we substitute the value of t into the equation "v = u + at":
v = 12 + (-9.8)(2.45)
v ≈ 12 - 24.01
v ≈ -12.01 m/s (negative sign indicates downward direction)
b. We have already calculated that the man is in the air for 2.45 seconds. This is the time from when he lets go of the rope until he lands on the pillow.
c. To find the greatest height reached by the man above the ground, we need to calculate his maximum displacement (or maximum height). We can use the equation for displacement again:
s = ut + (1/2)at^2
Since we know u (initial velocity) and t (time), we can substitute these values into the equation:
s = 12(2.45) + (1/2)(-9.8)(2.45)^2
Simplifying:
s = 29.4 - 30.1475
s ≈ -0.7475 m
Since the displacement is negative, it means the highest point the man reached is below his initial position. In this case, the pillow is situated 20 m below the man's initial position (as stated in the question). Therefore, the man's greatest height is above the ground and can be calculated by adding his displacement to the height of the pillow. So,
Greatest height reached = (-0.7475) + 20 ≈ 19.25 m
Therefore, the greatest height reached by the man above the ground is approximately 19.25 m.
d. To find the distance between the helicopter and the man three seconds after he lets go of the rope, we need to calculate the horizontal distance traveled by the helicopter in three seconds. Since the helicopter is moving directly upward and the horizontal velocity remains constant, the distance traveled will be equal to the horizontal velocity multiplied by the time taken.
The horizontal distance is given by:
Distance = (horizontal velocity) * time
= (0 m/s) * 3 s
= 0 m
Therefore, the distance between the helicopter and the man three seconds after he lets go of the rope is zero meters. This means the helicopter and the man are at the same horizontal position.