Different layers represent clouds made of gases that condense at different temperatures.
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
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:
To convert this into joules, we use the conversion factor:
So, 
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
I’m so sorry, I need more information. Good luck and I’m sorry I couldn’t help you :(
Answer:
a) 1.22 s
b) 9.089 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
Time taken by the ball to reach the highest point is 1.22 seconds
The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m
