What is the total distance, side to side, that the top of the building moves during such an oscillation? The New England Merchan
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!
Answer:
a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.
Explanation:
So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.
part a)
So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:
When solving for theta, we get that:
so now we can substitute the corresponding values:
Which yields:
but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:
θ=270°-60°=210°
part b)
for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:
The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:
part c)
In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:
and now up stream:
Once we got these two velocities we will now need to find the time to take each trip:
time down stream:
and the time up stream:
so the total time will be:
d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:
t=1.333hr
e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)
f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.
so
t= 1hr
