"Free fall" is the motion of an object when gravity is the ONLY force
acting on it.
In true 'free fall' the speed of an object increases at a constant rate
for the total duration of the fall. The rate of increase, on or near the
Earth's surface, is 9.8 meters per second for each second of fall.
True free fall is almost impossible to observe in everyday life, because
whenever we see anything falling, it's almost always falling through air,
so gravity is NOT the only force acting on it. The friction due to the
motion through air works against the gravitational force. In many cases,
the result is that the object's speed eventually stops increasing and
becomes constant, at a speed often described with the faux technical,
high-fallutin' sounding phrase "terminal velocity". It must be understood
that 'terminal velocity' is NOT a property of gravity or of free fall, but is
only a result of falling through some surrounding stuff that interferes with
the process of true 'free fall'.
Answer:
17.1130952381 s
No
Explanation:
t = Time taken
u = Initial velocity = 115 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6.72 m/s² (negative as it is decelerating)
From the equations of motion
The minimum time required to stop is 17.1130952381 s
The distance that is required for the jet to stop is 0.98400297619 km which is greater than 0.8 km. So, the jet cannot land on a small tropical island airport.
Answer:
a) r = 4.22 10⁷ m, b) v = 3.07 10³ m / s and c) a = 0.224 m / s²
Explanation:
a) For this exercise we will use Newton's second law where acceleration is centripetal and force is gravitational force
F = m a
a = v² / r
F = G m M / r²
G m M / r² = m v² / r
G M / r = v²
The squared velocity is a scalar and this value is constant, so let's use the uniform motion relationships
v = d / t
As the orbit is circular the distance is the length of the circle in 24 h time
d = 2π r
t = 24 h (3600 s / 1 h) = 86400 s
Let's replace
G M / r = (2π r / t)²
G M = 4 π² r³ / t²
r = ∛(G M t² / (4π²)
r = ∛( 6.67 10⁻¹¹ 5.98 10²⁴ 86400² / (4π²)) = ∛( 75.4 10²¹)
r = 4.22 10⁷ m
b) the speed module is
v = √G M / r
v = √(6.67 10⁻¹¹ 5.98 10²⁴/ 4.22 10⁷
v = 3.07 10³ m / s
c) the acceleration is
a = G M / r²
a = 6.67 10⁻¹¹ 5.98 10²⁴ / (4.22 10⁷)²
a = 0.224 m / s²
Answer:
1.18 sec
Explanation:
v = 0
u = 3.25m/s
v = u - gt
t1 = u/g = 3.25/9.8 = 0.332 s
height above board = v²/2g =
3.25²/(2×9.8) = 0.54 m
time to fall from there to water
t2 = √(2h/g) = √(2(3+ 0.54)/9.8) = 0.85 s
total time = t1 + t2
total time = 0.85 + 0.332 = 1.182 s
