On the report evaluation, the teacher wrote, "Good idea, but you
1 answer:
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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!
Let say for every 5 s of time interval the speed will remain constant
so it is given as
v(mi/h) 16 21 23 26 33 30 28
now we have to convert the speed into ft/s as it is given that 1 mi/h = 5280/3600 ft/s
so here we will have
v(ft/s) 23.5 30.8 33.73 38.13 48.4 44 41.1
now for each interval of 5 s we will have to find the distance cover for above interval of time
so here it will cover 1298.1 ft distance in 30 s interval of time
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
= m
+ m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s