A cannonball fired with an initial speed of 40 m/s and a launch angle of 30 degrees from a cliff that is 25m tall.
1 answer:
We know that the ball traveled with an initial velocity of 40m/s at a 30° angle above the horizontal. The image below shows how much of this velocity was upward velocity and how much was horizontal velocity. Upward velocity was 20 m/s and horizontal velocity was √(40) m/s, or 2√(10) m/s. We get these numbers from the ratios of the 30-60-90 triangle.
a) What is the flight time of the cannonball?
The flight time of the cannonball can be found by finding the time at which the upward velocity equals zero (the top of the ball's trajectory) and then finding how long it took to hit the ground after that point.
To find where upward velocity equals zero:
V = Vi - a(t) , where V equals vertical velocity, Vi equals initial vertical velocity, and a equals acceleration due to gravity (-9.8 m/s²)
V = 20 - 9.8(t) Set V equal to zero, because we want to find the moment when the ball reached the peak of its travel path
0 = 20 -9.8t Add 20 to both sides, then divide by 9.8
t = 2.041
This is the point where the ball was at the top of its trajectory.
At this point, How high was the ball?
d = Vi x t + (1/2) (a) (t²) , where d is distance traveled
d = 20(2.041) + (1/2) (-9.8) (2.041²)
d = 20.388
Remember that the ball was launched from 25 m above the ground, so add 25 to the height that the ball traveled from this point:
25 + 20.388 = 45.388
This was the height the ball reached before it started to come down. Plug this into the distance formula to see how long it took to hit the ground. Remember that this is similar to the ball being dropped from rest from this height, since vertical velocity was zero.
45.388 = (0)(t) - (1/2) (-9.8) (t²) Multiply both sides by (-2/-9.8)
9.26 = t²
t = 3.043
We know that it took 2.041 seconds to reach the peak height, and 3.043 seconds to come down.
Total flight time = 2.041 + 3.043 = 5.084 seconds
Remember that, neglecting air resistance, the ball will maintain the same horizontal velocity the entire time. This means the horizontal velocity was 10√2 during the entire flight time.
distance = velocity * time = 5.084 * 10√2 = 32.154
a) What is the flight time of the cannonball?
The flight time of the cannonball can be found by finding the time at which the upward velocity equals zero (the top of the ball's trajectory) and then finding how long it took to hit the ground after that point.
To find where upward velocity equals zero:
V = Vi - a(t) , where V equals vertical velocity, Vi equals initial vertical velocity, and a equals acceleration due to gravity (-9.8 m/s²)
V = 20 - 9.8(t) Set V equal to zero, because we want to find the moment when the ball reached the peak of its travel path
0 = 20 -9.8t Add 20 to both sides, then divide by 9.8
t = 2.041
This is the point where the ball was at the top of its trajectory.
At this point, How high was the ball?
d = Vi x t + (1/2) (a) (t²) , where d is distance traveled
d = 20(2.041) + (1/2) (-9.8) (2.041²)
d = 20.388
Remember that the ball was launched from 25 m above the ground, so add 25 to the height that the ball traveled from this point:
25 + 20.388 = 45.388
This was the height the ball reached before it started to come down. Plug this into the distance formula to see how long it took to hit the ground. Remember that this is similar to the ball being dropped from rest from this height, since vertical velocity was zero.
45.388 = (0)(t) - (1/2) (-9.8) (t²) Multiply both sides by (-2/-9.8)
9.26 = t²
t = 3.043
We know that it took 2.041 seconds to reach the peak height, and 3.043 seconds to come down.
Total flight time = 2.041 + 3.043 = 5.084 seconds
Remember that, neglecting air resistance, the ball will maintain the same horizontal velocity the entire time. This means the horizontal velocity was 10√2 during the entire flight time.
distance = velocity * time = 5.084 * 10√2 = 32.154
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Answer:
The Entropy generated by the steam = 2.821 kJ/K
Explanation:
Total volume of container = 5m³
Heat transfer does not exist between system and surrounding, dQ = 0
At the first chamber, temperature of water at saturated liquid is 300°C
From the steam table:
Specific enthalpy of saturated liquid at 300°C ,
Specific internal energy of saturated liquid at 300°C, 1332.7 kJ/kg
For closed system, the first law of thermodynamics state that:
dQ = dw + dU..................(1)
work done for free expansion, dw =0
0 = 0 + dU
dU = 0 , i.e. U₁ = U₂
At the second chamber,
The final pressure, P₂ = 50 kPa
From the steam table, at P₂ = 50 kPa, = 340.49 kJ/kg
Let the dryness fraction at the second chamber = x
Since U₁ = U₂
1332.7 = 340.49 + x2140.7
Dryness fraction, x = 0.463
From steam table, the specific volume is,
V₂ = 5 m³
1.5 = 5/m₂
m₂ = 3.33 kg
At 300°C
From the steam table,
Therefore the entropy generated will be :
Entropy = mass* (S₂ - S₁)
Entropy = 3.33* (4.102 - 3.2548)
Entropy = 2.821 kJ/K