Given that,
For glasses:
distance, s = 42 m
a= -g = -9.81 m/s²
we can calculate time taken by glasses to hit the ground as follow:
s= Vi* t + 1/2 at²
Since at top initial velocity, Vi= 0
42 = 0 - 0.5* 9.81 t²
t = 2.926 s
Pen is drop 2 second after the glasses is dropped. So time difference will be:
Δt = 2.926 - 2 = 0.926 s
Distance covered by Pen in 0.926 s can be calculated as:
s = 0 - 0.5* 9.81 * 0.926²
s= 4.2 m
Distance of pen from ground will be, 42-4.2 = 37.8 m
When the glasses hit the ground, the pen will be 37.8 m above the ground.
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Answer:
We know that for a pendulum of length L, the period (time for a complete swing) is defined as:
T = 2*pi*√(L/g)
where:
pi = 3.14
L = length of the pendulum
g = gravitational acceleration = 9.8 m/s^2
Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.
Then the period is independent of:
The mass of the child
The initial angle
Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.
Answer:
Explanation:
The vertical component of the initial velocities are
If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation
So the ratio of the times of the flights is