Assume the acceleration due to gravity is approximately 9.8 m/s^2
In 1 second, it will fall
s = ut + at^2/2
u = initial vertical velocity = 0
a = 9.8
t = 1
s is the distance it falls
= 9.8/2 = 4.9 m
But the parcel has a horizontal velocity of 1 m/s so it will travel 1 m horizontally in 1 second.
The answer would be
1 metre, 4.9 metres.
Answer:
a) v = 0, a = - 15 pi²
b) v = - 40.8 cm / s, a = -74.2 cm / s
Explanation:
the simple harmonic motion is described by the expression
x = A cos (wt + Ф)
in that case they indicate the amplitude of the movement
A = 15 cm
angular velocity and period are related
w = 2π / T
w = 2π/ 2
w = π rad / s
speed is defined by
v = dx / dt
v = -A w sin (wt + Ф)
acceleration is defined by
a = dv / dt
a = - A w² cos (wt + Ф)
In order to evaluate this expression we must know the value of the phase constant (Ф) that meets the initial conditions, the most common thing is that the system is released, v = 0 for y = 0 s, let's use the speed equation
v = -A w sin (0+ Ф)
for this expression to be zero, the sine must be zero, so
Ф= 0
now let's write the equations for our case
v = - 15 π sin (π t)
a = - 15 π² cos (π t)
a) for x = 0
v = 0
a = - 15 pi²
b) for x = 7.5 cm
Let's find out how long it takes to get to this position
x = 15 cos (π t)
7.5 / 15 = cos π t
pi t = cos⁻¹ 0.5
t = 1.047/π
don't forget that the angles are in radians
t = 0.333 s
now let's look for speed and acceleration
v = - 15 π sin (π 0.333)
v = - 40.8 cm / s
a = - 15 π² cos (π 0.333)
a = -74.2 cm / s
