Melissa and David were playing on the tire swing that hangs from the big oak tree in their yard. While Melissa was sitting in th
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Answer:
The angle for the forward Mach line is 19.47°
The angle for the rearward Mach line is 5.21°
Explanation:
From table A-1 (Modern Compressible Flow: with historical perspective):
(M₁ = 3)
If Po₁ = Po₂
Table A-1:
Table A-5:
v₁ = 49.76°
μ₁ = 19.47°
v₂ = 60.55°
μ₂ = 16°
θ = 60.55 - 49.76 = 10.79°
The angle for the forward Mach line is:
μ₁ = 19.47°
The angle for the rearward Mach line is:
θr = μ₂ - θ = 16 - 10.79 = 5.21°
Answer: a) The cliff is 532.05m high
b) Her speed just before hitting the ground is 102.12 m/s
Explanation: To solve This, I'll use a sketch diagram, attached to this solution,
In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,
Y1 = ut + 0.5g(t^2)
Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s
Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m
Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.
Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,
Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds
Y2 = 340 × 3 = 1020m
But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x
x = 487.95m
So, total height of cliff = 44.1 + 487.95 = 532.05m
b) the speed of the teacher just before she hits the ground.
Using the equations of motion again,
(V^2) = (U^2) + 2gs
Where v is the final velocity to be calculated
U is the initial velocity = 0m/s
g is acceleration due to gravity = 9.8m/s2
S is the total height she fell through, that is, the height of the cliff = 532.05m
(V^2) = 0 + 2×9.8×532.05 = 10428.18
V = √(10428.18) = 102.12m/s
QED!
