Answer:
The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s
Explanation:
Given,
Your velocity, V₁ = 2 m/
The velocity of the person, V₂ =?
The velocity of the person relative to you, V₂₁ = 3 m/s
According to the relative velocity of two
V₂₁ = V₂ -V₁
∴ V₂ = V₂₁ + V₁
On substitution
V₂ = 3 + 2
= 5 m/s
Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s
Temperature is how hot or cold something is,barometric is air pressure(when the pressures high,the weather is dry)humidity is how moist the air is and how many water particles are there, wind speed and direction is how the hot and cold air is moving and how fast,which makes wind (high to low, hot to cold) precipitation is how much it is raining at that point.
Between the stars' absolute magnitudes<span> or </span>luminosities<span> versus their </span>stellar classifications<span> or </span>effective temperatures<span>. </span>
They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
