Problem 38
A peregrine falcon dives at a pigeon. The falcon starts downward from rest and falls with free-fall acceleration. If the pigeon is 76.0-m below the initial position of the falcon, how long does it take the falcon to reach the pigeon? Assume the pigeon remains at rest.
SOLUTION:
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
t |
|
y = -76.0 m |
|
|
v0 = 0 |
|
y = y0 + v0t
– ½ gt2
-76.0 m = 0 + 0 – ½ (9.80 m/s2)t2
t2 = (-76.0 m)/( -4.90 m/s2)
t = 3.94 s
Problem 39
A small mailbag is released from a helicopter that is descending steadily at 1.50 m/s. After 2.00-s,
(a) What is the speed of the mailbag?
SOLUTION:
Note: Even though the bag was dropped from the helicopter, the initial velocity is NOT zero because the helicopter was moving. So what is the initial velocity of the bag? If we take down as negative, the helicopter was descending at 1.50 m/s, therefore the bag has an initial velocity of –1.50 m/s. We also take this drop point as our zero position.
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
v |
|
t = 2.00 s |
|
|
v0 =
-1.50 m/s |
|
v = v0 –
gt
v = -1.50 m/s – (9.80 m/s2)(2.00 s)
v = -21.1 m/s
(b) How far is it below the helicopter?
SOLUTION:
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
y |
|
t = 2.00 s |
|
|
v0 =
-1.50 m/s |
|
|
v = -21.1 m/s |
|
y = y0 + ½
(v + v0)t = 0 + ½ (-21.1 m/s – 1.50 m/s)(2.00 s)
y = -22.6 m
(c) Repeat (a) and (b) if the helicopter was rising steadily.
SOLUTION:
If the helicopter was rising, the only difference will be the velocity is positive or v0 = +1.50 m/s, therefore:
v
= v0 – gt
v = 1.50 m/s – (9.80 m/s2)(2.00 s)
v = -18.1 m/s
y = y0 + ½ (v + v0)t = 0 + ½ (-21.1 m/s + 1.50 m/s)(2.00
s)
y = -19.6 m
Problem 41
A ball thrown vertically upward is caught by the thrower after 2.00 s.
(a) Find the initial velocity of the ball.
SOLUTION:
If a ball is thrown from y0 = 0 then caught at the same spot, then y = 0, therefore:
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
v0 |
|
y = 0 |
|
|
t = 2.00 s |
|
y = y0 + v0t
– ½ gt2
0 = 0 + v0(2.00 s) – (4.90 m/s2)(2.00 s)2
(2.00 s)v0 = 19.6 m
v0 = 9.80 m/s
(b) Find the maximum height it reaches.
SOLUTION:
If it takes 2.00 s to return to its original position, then it takes half that, or 1.00 s, to reach maximum height. Also, the velocity at the maximum height is equal to ZERO.
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
y |
|
v0 =
9.80 m/s |
|
|
v = 0 |
|
|
t = 1.00 s |
|
y = y0 + ½
(v + v0)t = 0 + ½ (0 + 9.80 m/s)(1.00 s)
y = 4.90 m
Problem 42
A model rocket is launched straight upward with an initial speed of 50.0 m/s. It accelerates with a constant upward acceleration of 2.00 m/s2 until its engine stops at an altitude of 150 m.
(a) What is the maximum height reached by the rocket?
SOLUTION:
While the rocket is accelerating to 150 m height, it has a net acceleration acting on it of magnitude (9.80 – 2.00) = 7.80 m/s2 directed downward. Once the engine stops, the only acceleration left is due to gravity. We need to find the velocity of the rocket the moment the engine stopped in order to find its maximum height. We do this in two parts, first, the rocket with its engine on:
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
v |
|
y = 150 m |
|
|
v0 =
50.0 m/s |
|
|
a = -7.80 m/s2 |
|
v2 = (v0)2
+ 2a (y – y0)
v2 = (50.0 m/s)2 – 2 (7.80 m/s2)(150 m)
v2 = 160 m2/s2
v = 12.6 m/s
This velocity, v = 12.6 m/s, becomes the initial velocity for part 2 where we want to find the maximum height. Remember, at maximum height, v = 0.
|
KNOWNS |
UNKNOWN |
|
y0 = 150 |
y |
|
v = 0 |
|
|
v0 =
12.6 m/s |
|
v2 = (v0)2
– 2g (y – y0)
y = y0 + [(v2 – v02)/(-2g)]
y = 150 m + [-(12.6 m/s)2/(-19.6 m/s2)]
y = 150 m + 8.2 m = 158.2 m
(b) How long after lift-off does the rocket reach maximum height?
SOLUTION:
Once again, we do this in two parts just like part (a). First part, find the time till the engine shuts off.
|
KNOWNS |
UNKNOWN |
|
y0 = 0 |
t |
|
y = 150 m |
|
|
v0 =
50.0 m/s |
|
|
v = 12.6 m/s |
|
|
a = -7.80 m/s2 |
|
y = y0 + ½
(v + v0)t
150 m = 0 + ½ (50.0 m/s + 12.6 m/s)t
150 m = (31.3 m/s) t
t = (150 m)/(31.3 m/s) = 4.79 s
Then we do part 2 when the rocket engine shuts off till maximum height.
|
KNOWNS |
UNKNOWN |
|
y0 = 150 |
t |
|
v = 0 |
|
|
v0 =
12.6 m/s |
|
|
y = 158.2 m |
|
y = y0 + ½
(v + v0)t
158.2 m = 150 + ½ (0 + 12.6 m/s)t
8.2 m = (6.3 m/s) t
t = (8.2 m)/(12.6 m/s) = 1.29 s
The total time is the sum T =
4.79 s + 1.30 s= 6.07 s
(c) How long is the rocket in the air?
SOLUTION:
The rocket falls from a height of 158.2 m, how long does it take to hit the ground?
|
KNOWNS |
UNKNOWN |
|
y0 =
158.2 |
t |
|
y = 0 |
|
|
v0 = 0 |
|
y = y0 + v0t
– ½ gt2
0 = 158.2 m + 0 – (4.90 m/s2)t2
t = 5.68 s
Total time = 6.07 + 5.68 = 11.8 s